A child in free fall and an astronaut on the moon is will be weightless.
Weightlessness refers to the absence of weight, which is the gravitational force that an object exerts on another object. It occurs when an object is in a state of free fall.
Astronauts, when they're in space, experience weightlessness because they're in a state of free fall. It's the same experience that people would have if they were in an elevator and the cable snapped.
The moon's gravity is about one-sixth of the Earth's gravity. Therefore, an astronaut on the moon would weigh less than on Earth. Even though the astronaut wouldn't be completely weightless, he would be close enough to weightless that it would be hard to notice any difference in weight.
A child in the air as she jumps on a trampoline will also feel weightless when falling freely.
A scuba diver exploring a deep-sea wreck is not weightless. The force of gravity is still acting on the diver, pulling them downwards towards the seafloor. However, because the water provides an upward force called buoyancy, the diver may feel a sense of weightlessness or reduced weight compared to their weight on land. This is because the buoyant force counteracts some of the force of gravity acting on the diver, making them feel lighter. However, the diver still has mass and is not truly weightless.
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a lightbulb radiates most strongly at a wavelngth of abou t3000 nanometers. how hot is its filament?
The filament in a light bulb radiates light at a wavelength of 3000 nanometers, which corresponds to a temperature of 2700°C.
The temperature of a light bulb filament is directly related to the wavelength of the light it radiates.
The filament in a light bulb emits light at a wavelength of around 3000 nanometers, which is part of the visible light spectrum. This corresponds to a temperature of around 2700°C.
First understand the relationship between temperature and light emission.
As temperature increases, the wavelength of the emitted light decreases. This is known as Wien's law, and is expressed as:
λ = b/T
Where λ is the wavelength of the emitted light, b is a constant, and T is the temperature in Kelvin. As the temperature increases, the wavelength decreases.
The wavelength of 3000 nanometers (300 x 10^-9 m), the temperature of the filament must be around 2700°C.
This is very hot and is the reason why the filament can glow so brightly, producing the light that we use in our homes.
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for comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.1-ms-long encounter with a hard floor?
The magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.1-ms-long encounter with a hard floor is 9,819.819819819819 m/s².
Acceleration is defined as the rate at which velocity changes with time. Acceleration can be expressed as a vector with both magnitude and direction in physics. It's a scalar quantity in one dimension that only includes magnitude.
It is calculated as the ratio of the difference between the initial (v1) and final (v2) velocities of an object to the time interval (t) during which the velocity difference occurred. It's usually represented as:-
a = (v2 - v1) / t
The magnitude is the size of a vector or the scalar value of a physical quantity (that has a direction). Magnitude is used to describe how big an object or quantity is without taking its direction into account. The magnitude of the acceleration is the rate at which the speed of an object changes.
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a light has the frequency of 4.74 x 1014 sec-1 (hz). what is the wavelength? please show all the steps and all of your work when you upload your final answer.
The wavelength of a light wave can be calculated by the equation λ = c/f, where λ is the wavelength, c is the speed of light (3x[tex]10^{8}[/tex] m/s) and f is the frequency (4.74 x [tex]10^{14}[/tex] [tex]sec^{-1}[/tex]). Therefore, the wavelength of the light wave is 6.32 x [tex]10^{-7}[/tex] m.
The given frequency is 4.74 x 1014 sec-1. The formula to calculate the wavelength of a light wave is λ= c/f where c is the speed of light and f is the frequency of the light wave.
Therefore, λ= c/f= (3.00 x [tex]10^{8}[/tex] m/s)/(4.74 x [tex]10^{14}[/tex] [tex]sec^{-1}[/tex])= 6.32 x [tex]10^{-7}[/tex] m or 632 nm (rounding to three significant figures).
The wavelength of light is 6.32 x [tex]10^{-7}[/tex] m or 632 nm (rounding to three significant figures).
Formula to calculate the wavelength of a light wave: λ= c/f where c is the speed of light and f is the frequency of the light wave.
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: assuming equal mass, which will have the higher escape velocity from its surface, a large diameter planet or a small diameter planet?
Assuming equal mass, a small-diameter planet will have a higher escape velocity from its surface compared to a large-diameter planet.
This is due to the gravitational force being concentrated in a smaller area. The higher gravitational force from a smaller planet means that the escape velocity is greater, as the gravity is greater.
To calculate the escape velocity, we use the formula:
v = √(2GM/R), where G is the gravitational constant, M is the mass of the planet, and R is the radius.
We can see that the escape velocity is inversely proportional to the radius, so as the radius decreases, the escape velocity increases. This is why a small-diameter planet will have a higher escape velocity than a large-diameter planet with the same mass.
In conclusion, the escape velocity from the surface of a small-diameter planet will be higher than the escape velocity from the surface of a large-diameter planet, assuming they have the same mass.
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A banjo D string is 0.69 m long and has a fundamental frequency of 294 Hz.
Part A
Determine the speed of a wave or pulse on the string.
Express your answer to two significant figures and include the appropriate units.
v =
Part B
Identify first three other frequencies at which the string can vibrate.
Enter your answers using two significant figures in order of increasing frequencies separated by commas.
f2, f3, f4 =
Answer:
Part A:
The speed of a wave on the string can be calculated using the formula:
v = fλ
where f is the frequency and λ is the wavelength. In this case, we only know the frequency of the fundamental mode, so we need to use another formula that relates the wavelength and the length of the string:
λn = 2L/n
where n is the mode number (n = 1 for the fundamental mode), and λn is the wavelength of the nth mode. Substituting this expression for λ into the first formula, we get:
v = fn × 2L/n
Substituting the given values, we get:
v = (294 Hz) × 2(0.69 m)/(1)
v = 406 m/s
Therefore, the speed of a wave or pulse on the string is 406 m/s.
Part B:
The frequencies of the other modes of vibration can be calculated using the formula:
fn = nv/2L
where n is the mode number, v is the speed of the wave on the string (which we found in Part A), and L is the length of the string. Substituting the given values, we get:
f2 = (2 × 406 m/s)/(2 × 0.69 m) = 589 Hz
f3 = (3 × 406 m/s)/(2 × 0.69 m) = 883 Hz
f4 = (4 × 406 m/s)/(2 × 0.69 m) = 1178 Hz
Therefore, the first three other frequencies at which the string can vibrate are 589 Hz, 883 Hz, and 1178 Hz.
assuming unfirm radiation in all directions find he light intensity in eV/s*m2 at a distance of 1 m from he light source
Light intensity will be approximately 0.0796 eV/s*m2
The intensity of radiation (I) at a distance (r) from a point source of power (P) is given by the inverse-square law:
I = P / (4πr²)
where π is the mathematical constant pi (approximately 3.14159).
Assuming the radiation is in the form of photons with energy E, the light intensity in eV/s*m² can be calculated as follows:
Convert the power P into units of energy/time, where time is measured in seconds:
P = E × N,
where N is the number of photons emitted per second by the source.
Substitute P into the formula for intensity and solve for I:
I = E × N / (4πr²)
Express I in eV/s*m² by dividing by the elementary charge (e):
I (in eV/s*m²) = (E × N / e) / (4πr²)
Assuming a monochromatic source of light with energy E = 1 eV, the number of photons emitted per second N can be calculated from the power of the source using the formula:
N = P / E
Let's assume that the light source has a power of 1 watt (1 J/s), then N = 1 eV/s.
Substituting the values into the formula for intensity, we get:
I = (1 eV/s) / (4π × (1 m)²) = 0.0796 eV/s*m²
Therefore, the light intensity in eV/sm² at a distance of 1 m from a monochromatic source of light with energy E = 1 eV and power 1 watt is approximately 0.0796 eV/sm².
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Drag each label to the correct location on the chart.
Sort the statements based on whether the described outcomes result from thermal energy being added or being removed.
Particles move faster.
Particles move slower.
Temperature increases.
Temperature decreases.
Kinetic energy increases.
Kinetic energy decreases.
those are the options
The outcomes based on whether thermal energy is added or being removed are:
Thermal energy added :
Particles move fasterTemperature increasesKinetic energy increasesThermal energy being removed :
Particles move slower.Temperature decreases.Kinetic energy decreasesHow does thermal energy affect particles, temperature and kinetic energy ?When thermal energy is added to a substance, the particles absorb this energy and start moving faster, which means their kinetic energy increases. This leads to an increase in temperature because the faster-moving particles collide with each other more frequently, transferring this extra energy in the form of heat.
Therefore, an increase in temperature and an increase in the kinetic energy of particles result from thermal energy being added, while a decrease in temperature and a decrease in the kinetic energy of particles result from thermal energy being removed.
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your instructor challenges you and your friend to pull on the ends of a horizontal rope attached to a pair of scales in a tug-of-war, but in such a way that the scale readings on the scales are different. can this be done? explain.
Answer:
If the scale readings are different then there will be a net force on the person attached to the scales:
Consider any point on the rope - if the forces in each direction are the same there is no acceleration of the rope
F = Δm * a for any portion of the rope with mass Δm
If any portion of the rope is accelerated, the person attached to the rope must be accelerated
in young's singe slit experiment, if the width of the slit decreases, what happends to the width of the diffracted peaks?
In Young's single slit experiment, if the width of the slit decreases, the width of the diffracted peaks increases.
Young's experiment involves a single slit that diffracts light and produces a pattern of bright and dark fringes on a screen. The width of the slit affects the diffraction of light through the slit and determines the width of the bright fringes on the screen.
The narrower the slit, the greater the diffraction of light, which causes the bright fringes to become wider.
This is because diffraction causes the light waves to spread out as they pass through the narrow slit, leading to interference and the formation of bright and dark fringes on the screen.
Therefore, if the width of the slit decreases, the width of the diffracted peaks increases.
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we want to lift a load of 200 lb with an overhead system using pulleys that have an efficiency of 0.9. if we can provide a maximum input force of 103 lb, what is the minimum number of pulleys that we need?
We need at least one pulley to lift the load of 200 lb with an overhead system using pulleys that have an efficiency of 0.9, given that we can provide a maximum input force of 103 lb.
Assuming that the weight of the pulleys and the rope is negligible, we can use the formula,
Load = (Input Force / Efficiencies) ^ Number of Pulleys
where Load is the weight of the load we want to lift, Input Force is the force we apply to the system, Efficiency is the efficiency of each pulley, and Number of Pulleys is the number of pulleys we need.
Plugging in the given values,
200 lb = (103 lb / 0.9) ^ Number of Pulleys
Simplifying the equation,
Number of Pulleys = log (base 2) (200 / (103/0.9))
Number of Pulleys = log (base 2) (200 x 0.9 / 103)
Number of Pulleys = log (base 2) 1.983495
Number of Pulleys = 1
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calculate the kinetic energy of a ball of mass 50g travelling at 30m/s. how much work will need to bee done to stop the ball?
Taking into account the definition of kinetic energy and work, the kinetic energy of a ball of mass 50 g travelling at 30 m/s is 22.5 J and a work of 22.5 J must be done in an opposite direction to stop the ball.
Definition of kinetic energyKinetic energy is defined as the energy associated with bodies that are in motion.
Kinetic energy is defined as the amount of work necessary to accelerate a body of a certain mass and in a position of rest, until it reaches a certain speed. This will remain the same unless there is a change in speed or the body returns to its state of rest by applying a force.
Kinetic energy is represented by the following expression:
Ec = 1/2×m×v²
Where:
Ec is kinetic energy, which is measured in Joules (J).m is mass measured in kilograms (kg).v is velocity measured in meters over seconds (m/s).Definition of workThe kinetic energy theorem states that the work done by the applied net force (sum of all forces) is equal to the change in kinetic energy:
Work= Ec final - Ec original
Kinetic energy of the ballIn this case, you know:
m= 50 g= 0.05 kg (being 1000 g= 1 kg)v= 30 m/sReplacing in the definition of kinetic energy:
Ec = 1/2× 0.05 g× (30 m/s)²
Solving:
Ec= 22.5 J
The kinetic energy is 22.5 J.
Work to stop the ballStoping the ball means bringing the velocity to zero. This is:
Ec final= 1/2× 0.05 g× (0 m/s)²
Ec final= 0
Then, work can be calculated as:
Work= Ec final - Ec initial
Work= 0 - 22.5 J
Work= -22.5 J
This means that a work of 22.5 J must be done in an opposite direction.
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calculate the frequency, in megahertz, of the accelerating voltage needed for a proton in a 1.15-t field.
The frequency of the accelerating voltage needed for a proton in a 1.15-t field is 28.1 MHz.
The cyclotron frequency is
f = qB/2πm
where f is the frequency in Hertz (Hz),
q is the charge of the proton in Coulombs,
B is the magnetic field strength in Tesla, and
m is the mass of the proton in kilograms.
For a proton, the charge is q = 1.602*10⁻¹⁹ C,
and the mass is m = 1.673*10⁻²⁷ kg.
If the magnetic field strength is given as B = 1.15 T, then we can plug in the values into the formula and calculate the frequency:
f = (1.602*10⁻¹⁹ C)(1.15 T)/(2π)(1.673*10⁻²⁷kg)
= 28.1 MHz
Therefore, the frequency of the accelerating voltage needed for a proton in a 1.15 T field is approximately 28.1 MHz.
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determine if the drag force exerted on an object moving through air (a.k.a. force of air resistance) is proportional to the velocity or the square of the velocity of the object.
The drag force exerted on an object moving through air (a.k.a. force of air resistance) is proportional to the square of the velocity of the object.
Thus, the correct answer is proportional to the square of the velocity of the object.
What is the drag force?The аir resistаnce force аcting on аn object moving through аir is referred to аs drаg force. When а body trаvels through а fluid such аs wаter or аir, it fаces resistаnce to its motion, which is proportionаl to the velocity of the object. This resistаnce force аcting on а body moving through аir is referred to аs аir resistаnce or drаg force.
The drаg force on аn object in the аir is proportionаl to the squаre of the object's velocity. When the velocity of the object is doubled, the drаg force becomes four times greаter. Thus, the drаg force grows fаster thаn the object's velocity. In other words, the drаg force аcting on аn object increаses аs the squаre of the object's velocity.
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Part 1: An open vertical tube has water in it. A tuning fork vibrates over its mouth. As the water level is lowered in the tube, the seventh resonance is heard when the water level is 198.25 cm below the top of the tube. 198.25 cm What is the wavelength of the sound wave? The speed of sound in air is 343 m/s. Answer in units of cm.
Part 2: What is the frequency of the sound wave; i.e., the tuning fork? Answer in units of s^−1 .
Part 3: The water continues to leak out the bottom of the tube. When the tube next resonates with the tuning fork, what is the length of the air column? Answer in units of cm.
Part 1: The wavelength of the sound wave is 47.26 cm
Part 2: Frequency is 725.49 s^-1
Part 3: The length of the air column when the tube resonates again is 70.695 cm.
How did we get these values?Part 1:
The wavelength of the sound wave can be calculated using the formula:
wavelength = 4L/n
where L is the length of the air column in the tube and n is the harmonic number (in this case, n = 7).
Given that the water level is 198.25 cm below the top of the tube, the length of the air column is:
L = total length of tube - water level = 4L - 198.25
Solving for L, we get:
L = 198.25/3 = 66.083 cm
Therefore, the wavelength of the sound wave is:
wavelength = 4L/n = 4(66.083)/7 = 47.26 cm
Part 2:
The speed of sound in air is 343 m/s. The frequency of the sound wave can be calculated using the formula:
frequency = speed of sound / wavelength
Substituting the values we have:
frequency = 343 / (47.26/100) = 725.49 s^-1
Part 3:
When the tube resonates again, the air column length will be equal to a multiple of half-wavelengths. Since the tube was already in its 7th harmonic, the next resonance will occur in the 8th harmonic, which means the air column will be equal to 8 times half-wavelengths.
So, the length of the air column can be calculated using the formula:
L = (2n-1)wavelength/4
where n is the harmonic number (in this case, n = 8).
Substituting the values we have:
L = (2(8)-1)(47.26/2)/4 = 282.78/4 = 70.695 cm
Therefore, the length of the air column when the tube resonates again is 70.695 cm.
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determine the capacitance of a teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.020 0 mm.
To determine the capacitance of a teflon-filled parallel-plate capacitor having a plate area of 1.80[tex]cm^{2}[/tex] and a plate separation of 0.0200 mm, we can use the formula for capacitance: C = εo εr A/d, when the values are plugged in, the capacitance is found to be [tex]1.54* 10^{-9}[/tex] Farads.
The capacitance of a teflon-filled parallel-plate capacitor having a plate area of 1.80[tex]cm^{2}[/tex] and a plate separation of 0.0200 mm is determined using the formula C = εo A/d, where C is the capacitance, εo is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
To explain this calculation further, the permittivity of free space is a constant value equal to [tex]8.85 * 10^{-12}[/tex] A/d, which is derived from the equation εo = 1/ (μoc2), where μo is the permeability of free space, and c is the speed of light. The area of the plates is given in the problem statement as 1.80 [tex]cm^{2}[/tex], and the distance between the plates is given as 0.0200 mm.
When these values are plugged into the formula, the capacitance is found to be [tex]1.54* 10^{-9}[/tex]Farads. In conclusion, the capacitance of a teflon-filled parallel-plate capacitor having a plate area of 1.80 [tex]cm^{2}[/tex] and a plate separation of 0.0200 mm is 1.54 x 10-9 Farads.
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a stone is dropped into a well. the sound of the splash is heard 3.00 s later. what is the depth of the well?
A stone is dropped into a well. the sound of the splash is heard 3.00 s later. The depth of the well is: 510 m
A stone is dropped into a well and the sound of the splash is heard 3.00 s later. To calculate the depth of the well, we can use the equation :
Depth = (Speed of sound x Time taken)/2
where the Speed of sound is 340 m/s. Therefore, the depth of the well is calculated to be 510 m.
To explain this in more detail, the equation states that the depth of the well is calculated by multiplying the speed of sound by the time taken for the sound to reach the surface of the well. This is then divided by two as the sound wave needs to travel to the bottom of the well and then back up to the surface.
In this case, the speed of sound is 340 m/s and the time taken for the sound to reach the surface is 3.00 s, so the depth of the well is 510 m.
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a force of 1150 N acts parallel to ramp to push a 250 kg gun safe into a moving van. The ramp is frictionless and inclined at 17 degree
a) what is the acceleration of the safe up the ramp?
b) If we consider friction in this problem, with a friction force of 120 N, what is the acceleration of the safe
A force of 1150 N acts parallel to ramp to push a 250 kg gun safe into a moving van. The ramp is frictionless and inclined at 17 degree
a) To find the acceleration of the safe up the ramp,
1. Determine the component of the force parallel to the ramp: F_parallel = 1150 N
2. Calculate the gravitational force acting on the safe along the ramp: F_ gravity = m * g * sin(theta) = 250 kg * 9.81 m/s^2 * sin(17 degrees) ≈ 729.6 N
3. Calculate the net force acting on the safe: F_ net = F_ parallel - F_ gravity = 1150 N - 729.6 N ≈ 420.4 N
4. Use Newton's second law to find the acceleration: F_ net = m * a, so a = F_ net / m = 420.4 N / 250 kg ≈ 1.68 m/s^2
The acceleration of the safe up the ramp is approximately 1.68 m/s^2.
b) To find the acceleration of the safe considering friction, follow these steps:
1. Determine the friction force: F_friction = 120 N
2. Calculate the net force acting on the safe, considering friction: F_net_with_friction = F_parallel - F_gravity - F_friction = 1150 N - 729.6 N - 120 N ≈ 300.4 N
3. Use Newton's second law to find the acceleration with friction: a_with_friction = F_net_with_friction / m = 300.4 N / 250 kg ≈ 1.20 m/s^2
When considering friction, the acceleration of the safe up the ramp is approximately 1.20 m/s^2.
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what if? for the value of e found in part (a), what would the kinetic energy of a proton have to be (in mev) for it to move undeflected in the negative x-direction?
Answer: The kinetic energy of the proton would have to be 0.0209 MeV for it to move undeflected in the negative x-direction.
For the given problem, if the value of e found in part (a), the kinetic energy of a proton would have to be 4.31 MeV for it to move undeflected in the negative x-direction. The solution to this problem is given below: Given information:
Electric field = 1.1 kV/m
Proton mass = 1.67 x 10-27 kg
Charge of proton = 1.6 x 10-19 C
Taking the given data, the equation of motion for a proton with an initial velocity at right angles to the electric field is given by the equation, F = qE
Here, F is the electric force on the proton, q is the charge of the proton and E is the electric field strength. If a magnetic field is also present, then a proton will also be subject to the Lorentz force, F = qvB where v is the velocity of the proton and B is the magnetic field strength.
Then, the equation of motion for a proton moving at a speed v in a uniform magnetic field is given by the equation,
F = q(vB sin θ) (1)
Where θ is the angle between the direction of motion of the proton and the direction of the magnetic field.
The speed v of the proton when moving undeflected is given by the equation,
F = qE (2)
Combining the above equations, we get,
qE = q(vB sin θ) (3)Here, the value of θ is 90 degrees because the proton is moving perpendicular to the magnetic field. Thus, sin θ = 1. So, the equation (3) becomes,v = E/B = 1.1 x 103 / 0.55 = 2000 m/s
Now, the kinetic energy of the proton is given by the equation, K = 1/2mv2where m is the mass of the proton and v is its velocity.
So, putting the values of m and v, we get,
K = (1/2)(1.67 x 10-27)(2000)2 = 3.34 x 10-21 J
This is the kinetic energy of the proton when it is moving undeflected in the negative x-direction. We can convert this value into MeV by dividing it by 1.6 x 10-13 J/MeV.
Kinetic energy of the proton = 3.34 x 10-21 J= (3.34 x 10-21) / (1.6 x 10-13) = 0.0209 MeV
So, the kinetic energy of the proton would have to be 0.0209 MeV for it to move undeflected in the negative x-direction.
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a 0.60 kg block on a surface of negligible friction is pulled by a string which is passed over a pulley of negligible mass and friction, and is connected to a hanging 0.20 kg block. in terms of acceleration due to gravity g
g/2 is the acceleration
Let the tension in the string pulling 0.60 kg block is T
In a pulley system tension will be the same throughout the string
for 0.60kg block:
mg-T = ma
0.60g-T = 0.60a ..............(1)
for 0.20kg block:
T-mg = ma
T - 0.20g = 0.20a .............(2)
Solving equation 1 and 2:
(1)+(2)
0.60g-0.20g = 0.60a+0.20a
a = (0.60-0.20)g/(0.60+0.20)
a = 0.40g/0.80
a = g/2
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al is floating freely in his spacecraft, and you are accelerating away from him with an acceleration of 9.8 m/s2. how will you feel in your spacecraft? group of answer choices you will feel weight, but less than on earth. you will feel yourself pressed against the back of your spaceship with great force, making it difficult to move. you will feel the same weight as you do on earth. you will be floating weightlessly. you will feel weight, but more than on earth.
You will feel yourself pressed against the back of your spacecraft with great force, making it difficult to move. This is because when you accelerate, the force of gravity is increased, causing you to feel an increased weight.
This option is the correct one. As per Newton's second law, the force of the body is directly proportional to the mass and acceleration. Here, the acceleration is 9.8 m/s2, which means you will feel yourself pressed against the back of your spaceship with great force, making it difficult to move, you will feel the same weight as you do on earth.
This is an incorrect option because the acceleration is greater than 1g, which means the weight will be greater than the actual weight.you will be floating weightlessly. This is an incorrect option because there is an acceleration, which means you will not be floating weightlessly.you will feel weight, but more than on earth. This is an incorrect option because the acceleration is greater than 1g, which means the weight will be greater than the actual weight.
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According to the formula E=mc2:a. mass has to travel at the speed of light before it can produce any energyb. energy can travel much faster than light (in fact its speed can be the speed of light squared)c. a little bit of mass can be converted into a substantial amount of energyd. when two masses collide, we always get a lot of lighte. Einstein was a male chauvinist twice over
Answer:conversion factor between time and distance. If you have a certain amount of time tt, you can calculate the corresponding amount of distance by multiplying it by cc.Note: I'm not talking about the distance any particular object travels in the time tt. If you have a car traveling at speed vv, you can find out how far the car moves in time tt by multiplying by vv, but that's not converting time into distance. The conversion is something more fundamental.The fact that time and distance can be converted into each other like this is one of the ways relativity changed our view of the world. One of its consequences is that speeds can now be measured as pure, unitless numbers. How so? Well, normally, we might measure distance in meters and time in seconds. So when you calculate a velocity as distance divided by time, you get an answer in meters per second. But because time can be converted into distance, now you can measure time in meters as well. So if you divide distance (in meters) by time (in meters), the meters cancel out and you get a pure number.As a pure number, c=1c=1. There are a few things that travel at speed 1, including light and gravity. Light was simply the first one that we discovered, which is the only reason cc is called the "speed of light."Once you see that cc is important for reasons having nothing to do with the fact that light travels at that speed, hopefully it seems less strange that it enters into the formula E=mc2E=mc2. Just as you can convert time into distance, you can also convert mass into energy. You just have to multiply by cc twice, not just once, to make the units work out.
Explanation:conversion factor between time and distance. If you have a certain amount of time tt, you can calculate the corresponding amount of distance by multiplying it by cc.Note: I'm not talking about the distance any particular object travels in the time tt. If you have a car traveling at speed vv, you can find out how far the car moves in time tt by multiplying by vv, but that's not converting time into distance. The conversion is something more fundamental.The fact that time and distance can be converted into each other like this is one of the ways relativity changed our view of the world. One of its consequences is that speeds can now be measured as pure, unitless numbers. How so? Well, normally, we might measure distance in meters and time in seconds. So when you calculate a velocity as distance divided by time, you get an answer in meters per second. But because time can be converted into distance, now you can measure time in meters as well. So if you divide distance (in meters) by time (in meters), the meters cancel out and you get a pure number.As a pure number, c=1c=1. There are a few things that travel at speed 1, including light and gravity. Light was simply the first one that we discovered, which is the only reason cc is called the "speed of light."Once you see that cc is important for reasons having nothing to do with the fact that light travels at that speed, hopefully it seems less strange that it enters into the formula E=mc2E=mc2. Just as you can convert time into distance, you can also convert mass into energy. You just have to multiply by cc twice, not just once, to make the units work out.
Worked Calculate the number of electrons that a positively charged object gains if its charge decreases by 3,2 x 10-18 C.
The positively charged object gains 20 electrons when its charge decreases by 3.2 x 10^-18 C.
What is Positive Charge?
A positive charge is an electrical property of matter that describes the presence of more positively charged protons than negatively charged electrons in an atom or molecule. In other words, an object with a positive charge has lost one or more electrons, resulting in a net charge that is greater than zero.
We know that the charge on a single electron is 1.602 x 10^-19 C.
To calculate the number of electrons gained by a positively charged object when its charge decreases by 3.2 x 10^-18 C, we can use the formula:
number of electrons = (magnitude of charge lost) / (charge on a single electron)
number of electrons = (3.2 x 10^-18 C) / (1.602 x 10^-19 C)
number of electrons = 20
Therefore, the positively charged object gains 20 electrons when its charge decreases by 3.2 x 10^-18 C.
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if a wavelength is 635 nm, what is the frequency? please show all the steps and all of your work when you upload your final answer.
If a wavelength is 635 nm, the frequency is 4.72 × 10¹⁴ Hz.
The frequency of a wavelength is determined by the formula f = c/λ, where f is the frequency, c is the speed of light (3.00 x 108 m/s), and λ is the wavelength.
Given,
Wavelength = 635 nm
To find, frequency
Formula
The velocity of light = Wavelength × Frequency.
C = λ × f
Frequency f = C / λ
Where C = 3 × 10⁸ m/s, λ = 635 nm = 635 × 10⁻⁹ m
∴ f = C / λ
= (3 × 10⁸ m/s) / (635 × 10⁻⁹ m)
= (3 × 10⁸) × (10⁹ / 635)Hz= 4.72 × 10¹⁴ Hz
Frequency = 4.72 × 10¹⁴ Hz
Therefore, the frequency is 4.72 × 10¹⁴ Hz.
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determine how much the path of the bullet is altered by the magnetic field after it has traveled 1500 m
The path of the bullet is altered by the magnetic field after it has traveled 1500 m.
Magnetic field is an area in which an object experiences a force because of being placed within it. When the bullet is moved through a magnetic field, it experiences a force that alters its course from a straight line.
The force that a moving charged particle experiences as it travels through a magnetic field is known as the magnetic force.
Because the bullet is moving, the charge in the bullet is experiencing a magnetic force as a result of the magnetic field.The path of the bullet is altered by the magnetic field after it has traveled 1500 m.
The path of the bullet may be altered in various ways depending on the strength of the magnetic field and the velocity of the bullet.
When a magnetic field is generated in the vicinity of the bullet, the bullet's trajectory is modified. The bullet will always curve to the right or left, depending on the direction of the magnetic field.
The force acting on a charge traveling through a magnetic field is proportional to the speed of the charge, the charge on the charge, and the strength of the magnetic field.
Therefore, if a bullet is traveling at a high speed and encounters a strong magnetic field, it will be deflected in a sharp curve, resulting in a significant path deviation.
The path deviation of a bullet moving through a magnetic field after it has traveled 1500 m can be determined using the equation:f = (qVB)/m
where f is the force on the charged particle, q is the charge on the particle, V is the velocity of the particle, B is the magnetic field strength, and m is the mass of the charged particle.
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a flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. the gasoline burned in a 305-mile trip in a typical midsize car produces about 1.93 x 10^9 j of energy. how fast would a 25.5-kg flywheel with a radius of 0.284 m have to rotate to store this much energy? give your answer in rev/min.
The flywheel would need to rotate at a rate of 725 rev/min to store the given energy.
The rotational kinetic energy of a flywheel is given by the equation:
Ek = 1/2Iω²
where I is the moment of inertia and ω is the angular velocity.
The moment of inertia of a solid disk is given by: I = mr², where m is the mass and r is the radius of the disk.
Thus, substituting the given values, we have:
Ek = 1.93 x 10⁹ J.
Ek = 1/2 * (25.5 kg * (0.284 m)²) * ω²
1.93 x 10⁹ J = 1/2 * (25.5 kg * (0.284 m)²) * ω²
1.93 x 10⁹ J = 102 x 10⁻² ω²
ω² = 1.93 x 10⁹/102 x 10⁻²
ω² = 0.018 x 10⁷
ω² = 18 x 10⁴
ω = √18 x 10⁴
ω = 76 x 10² rad/s.
ω = 7600 rad/s.
Solving for ω, we get ω = 7600 rad/s.
We can convert this to rev/min by dividing by (2*pi) and multiplying by 60, giving us: ω = 725 rev/min.
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The human ear canal is about 2.6 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fundamental frequency around which we would expect hearing to be most sensitive? Assume the speed of sound in air to be 338 m/s. Answer in units of kHz
6.85 kHz is the basic frequency that we would anticipate having the greatest sensitivity in hearing.
A periodic waveform's lowest frequency is referred to as the fundamental frequency, or just the fundamental. The fundamental frequency is determined by the length of the tube, which in this case is 2.6 cm. The equation for the fundamental frequency of a tube open at one end and closed at the other is [tex]f=\frac{v}{2L}[/tex],
where L is the tube's length and v is the sound-traveling speed in air. In this situation
we have
[tex]f=\frac{338 m/s}{2(0.026 m)}\\\\f = 6.85 kHz.[/tex]
Consequently, we would anticipate that hearing would be most sensitive around the fundamental frequency of 6.85 kHz.
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two blocks are connected by a massless string that passes over a massless pulley. in the absence of friction, how does the tension force exerted by the string on the 250 n block compare with the tension force exerted by the string on the 350 block
Since the string is massless and the pulley is massless and frictionless, the tension force in the string is the same on both sides of the pulley. This means that the tension force exerted by the string on the 250 N block is the same as the tension force exerted by the string on the 350 N block.
This can be explained by considering Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force on each block is equal to the tension force in the string, since there is no friction. Since the blocks are connected by the string and the pulley, they both have the same acceleration. Therefore, the net force on each block must be the same.
Thus, the tension force exerted by the string on the 250 N block is equal to the tension force exerted by the string on the 350 N block.
The tension force exerted by the string on the 250 N block is equal to the tension force exerted by the string on the 350 N block. This is because, in the absence of friction, the forces acting on the blocks are balanced and in equilibrium.
When the 250 N block is pulled down, the 350 N block is pulled up with the same magnitude of force. This is due to the Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Thus, the tension force exerted by the string on the 250 N block is the same as the tension force exerted by the string on the 350 N block. This is the case regardless of the masses of the blocks, since the string and pulley are massless. Therefore, tension forces on both the blocks are equal.
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two charges that are 1 meter apart repel each other with a force of 2 n. if the distance between the charges is increased to 2 meters, the force of repulsion will be
Answer: Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 meters, the force of repulsion will be 0.5 N.
What is Coulomb's law?
Coulomb's law is a scientific law that relates to the interaction between two electrically charged objects. The power of the force acting between two point charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.
Coulomb's law equation can be written as: F = k(q₁q₂)/r²
Where, F is the electric force, q₁ and q₂ are charges, r is the distance between charges, and k is a constant with a value of 8.99 x 10⁹ Nm²/C².
So, when the distance between two charges is doubled, the force of repulsion decreases by a factor of four (2²).
Therefore, if the distance between the two charges is increased to 2 meters, the force of repulsion will be 2/4 = 0.5 N.
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one bullet is fired horizontally and a second bullet is simultaneously dropped from the same height. ignoring air resistance, which bullet will hit the ground first?
Both bullets will hit the ground at the same time, regardless of their initial horizontal velocity or any other factors, as long as air resistance is negligible. This is because, in the absence of air resistance, the horizontal motion of the fired bullet does not affect the time it takes to fall to the ground.
When the two bullets are released at the same height, they both have the same initial vertical velocity of zero. Therefore, they will both experience the same acceleration due to gravity as they fall toward the ground, and reach the ground at the same time. This phenomenon is famously demonstrated by Galileo's experiment of dropping objects of different masses from the Leaning Tower of Pisa. Despite the different masses, they all hit the ground at the same time because they experience the same acceleration due to gravity.
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a 10 gauge copper wire carries a current of 21 a. assuming one free electron per copper atom, calculate the magnitude of the drift velocity of the electrons.
To calculate the magnitude of the drift velocity of the electrons ,
The drift velocity of electrons in a conductor is given by the formula:
v = I / (neA)
where 'v' is the drift velocity of electrons,
'I' is the current flowing through the wire,
'n' is the number of free electrons per unit volume,
'e' is the charge on each electron, and
'A' is the cross-sectional area of the wire.
Therefore, The current-carrying capacity of the 10 gauge copper wire is
I = 21 A which is a given statement.
For copper, the number of free electrons per unit volume is approximately [tex]8.5*10[/tex]²⁸ electrons/m³, and the charge on each electron is 1.6 x 10⁻¹⁹ C.
The cross-sectional area of a 10 gauge copper wire is approximately 5.26 mm²= 5.26 x 10⁻⁷ m².
Substituting these values into the formula of drift velocity we get:
v = (21 A) / ((8.5 x 10²⁸ electrons/m³) x (1.6 x 10⁻¹⁹ C/electron) x (5.26 x 10⁻⁷ m²))
= 0.015 m/s
Therefore, the magnitude of the drift velocity of the electrons in the wire is approximately 0.015 m/s.
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