Albert Einstein was a German-born theoretical physicist who developed the theory of general relativity, effecting a revolution in physics.
Where is the contradiction between quantum physics and Einstein’s gravity?Rμν−12gμνR=8πGT^μν.
This is Einstein’s field equation. Essentially, this equation is general relativity. The left-hand side represents the geometry of spacetime. The right-hand side, the energy, momentum, and stresses of matter.
What this equation describes, in the words of Wheeler, is this: Spacetime tells matter how to move; matter tells spacetime how to curve.
But look closely. That T
on the right-hand side. It has a hat.
It has a hat because it is a quantum-mechanical operator. Because we know that matter consists of quantum fields. So it is described by operator-valued quantities (Dirac called them q-numbers). They are unlike ordinary numbers. For instance, when you multiply them, the order in which they appear matters. That is, when you have two operators p^
and q^
, p^q^≠q^p^
most of the time. So they are definitely not like numbers.
When Einstein wrote down his field equation over 100 years ago, the T
did not have a hat. But that’s because they didn’t know about operator-valued quantities at the time. Now we do. So I have to put the hat there.
But there are no hats on the left-hand side. And because of that, my equation might as well say something like, some apples = some oranges. It makes no sense. The stuff on the left-hand side (which consists of numbers) can never equal the stuff on the right-hand side (which definitely does not consist of numbers.)
I can make it work, though. I can replace that operator with its so-called expectation value:
Rμν−12gμνR=8πG⟨Tμν⟩.
This is called semiclassical gravity. And it works well, very well indeed. A little too well, as a matter of fact. Gravity is so weak, quantum effects are so irrelevant, this equation accurately describes Nature everywhere we can look. But we still don’t like it, because using that expectation value trick is a cheat, a cop-out.
Now you might wonder, why don’t I put hats on top of the things on the left-hand side? I would… if I knew how to quantize spacetime. That is, how to turn the numbers that describe gravity into quantum-mechanical operators.
But I do not. And nobody does. The standard methods all fail, leading to equations that make no sense at all.
So we are kind of stuck… we don’t know how to quantize gravity, and our observations don’t help us, don’t offer any hints as to how to get beyond semiclassical gravity. Theorists keep trying to come up with new ideas (or recycle old ones) but basically, we’ve been pretty much just spinning our wheels for decades.
Two asteroids are suspended in space 50 meters apart. The masses of the asteroids are 2000000 kg and
3000000 kg.
Answer:
Explanation:
What is the gravitational force between them?
To calculate the gravitational force between two objects, we can use the formula:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (6.6743 x 10^-11 N * m^2 / kg^2), m1 and m2 are the masses of the two objects, and r is the distance between them.
Plugging in the given values, we get:
F = (6.6743 x 10^-11 N * m^2 / kg^2) * (2000000 kg) * (3000000 kg) / (50 m)^2
F = 0.8046 N
Therefore, the gravitational force between the two asteroids is approximately 0.8046 N.
Which material would you choose to make the handrails of the playhouse? Use the data to explain your reasoning.
Answer:
Explanation:
To choose a material for the handrails of the playhouse, we need to consider its strength and durability. One option could be stainless steel, which has a high tensile strength and is resistant to corrosion and weathering. Another option could be treated wood, which is also strong and can be treated to resist moisture and insects. Ultimately, the choice would depend on factors such as cost, aesthetics, and availability of materials.
The driver of a car with a total of 1800 kg mass is traveling at 23 m/s when he slams on the brakes, locking the wheels on the dry pavement. The coefficient of kinetic friction between rubber and dry concrete is typically 0.7. How far would the car travel if were going twice as fast
Answer:
To solve this problem, we can use the formula:
d = (v^2)/(2μg)
d = distance traveled
v = speed of the car
μ = coefficient of kinetic friction
g = acceleration due to gravity
First, let's calculate the distance traveled when the car is traveling at 23 m/s:
d = (23^2)/(2*0.7*9.81) ≈ 67.97 meters
Now, let's calculate the distance traveled when the car is going twice as fast (46 m/s):
d = (46^2)/(2*0.7*9.81) ≈ 271.88 meters
Therefore, the car would travel approximately 271.88 meters if it were going twice as fast.
why the ocean near Christchurch is a different temperature than we’d expect for its latitude
Why the ocean near Christchurch is a different temperature than we'd expect for its latitude (distance from the equator)? Water moving from the equator is warmer than would be expected based on latitude, and so is warmer than the air it passes.
Changes to prevailing winds affect ocean currents. Changes to ocean currents affect how much energy is brought to (or taken away from) a location. In El Niño years, the prevailing winds that normally drive a warm current from the Equator past New Zealand are disrupted and may stop or even reverse.
A 1.20 kg copper rod resting on two horizontal rails 0.90 m apart carries a
current I = 55.0 A from one rail to the other. The coefficient of static friction
between the rod and rails is μs= 0.60.
(a) What is the smallest vertical magnetic field B that would cause the rod to
slide?
(b) Suppose a B field is directed at some angle to the vertical φ, with the current
along the rod directed into the page, as shown. Find an expression for B as a
function of φ for the case when the rod is just on the verge of beginning to slide.
(c) Find the value of φ which yields the smallest value of B that would cause
the rod to slide, together with the corresponding value of B.
Answer:
Explanation:
(a) In order for the copper rod to slide, the magnetic force on it must be greater than the maximum static friction force. The magnetic force on the rod can be found using the formula F = BIL, where B is the magnetic field, I is the current, and L is the length of the rod. The maximum static friction force can be found using the formula Ff = μsN, where μs is the coefficient of static friction and N is the normal force on the rod.
Since the rod is resting on two rails, the normal force on the rod is equal to its weight, N = mg, where g is the acceleration due to gravity. Therefore, the condition for the rod to slide is:
BIL > μs mg
Solving for B, we get:
B > μs mg / IL
Substituting the given values, we get:
B > (0.60)(1.20 kg)(9.81 m/s^2) / (0.90 m)(55.0 A)
B > 0.077 T
Therefore, the smallest vertical magnetic field that would cause the rod to slide is 0.077 T.
(b) When the rod is on the verge of beginning to slide, the magnetic force on it is equal to the maximum static friction force, F = Ff = μsN. The magnetic force can be expressed as F = BIL, and the normal force can be expressed as N = mg. Therefore, we have:
BIL = μs mg
Solving for B, we get:
B = μs mg / IL
But we also know that the angle between the magnetic field and the vertical is given by φ, so we can express L in terms of φ using the formula L = d/sinφ, where d is the distance between the rails. Therefore, we have:
B = μs mg sinφ / Id
(c) To find the value of φ that yields the smallest value of B, we need to minimize the expression for B with respect to φ. Taking the derivative of B with respect to φ, we get:
dB/dφ = μs mg cosφ / Id sin^2φ
Setting this derivative equal to zero and solving for φ, we get:
tanφ = μs mg / Id
Substituting the given values, we get:
tanφ = (0.60)(1.20 kg)(9.81 m/s^2) / (0.90 m)(3000000 kg)(9.00 x 10^-7 m)
tanφ ≈ 0.12
Taking the arctan of both sides, we get:
φ ≈ 6.87°
Substituting this value of φ back into the expression for B, we get:
B = μs mg sinφ / Id
B ≈ 0.030 T
Therefore, the smallest value of B that would cause the rod to slide is approximately 0.030 T, when the magnetic field is at an angle of 6.87° to the vertical.
Before a collision, a 200-kg Honda is driving 30 m/s towards a
600-kg Toyota that is not moving. After the crash, the two cars
are stuck together. What is their velocity?
m/s
As a result, the combined Honda-Toyota system's post-collision speed is 7.5 m/s.
How can you calculate the entire momentum prior to a collision?The system's center of mass was v/2 before to the collision since one automobile had a velocity of v and the other zero. The total momentum is equal to the entire mass times the velocity of the center of mass, or (2m)(v/2) = mv before and after.
Initial momentum of Honda = m1 * v1
= 200 kg * 30 m/s
= 6000 kg·m/s
Final momentum of combined system = (m1 + m2) * v_final
Setting the two momenta equal to each other, we get:
6000 kg·m/s = 800 kg * v_final
Solving for v_final, we get:
v_final = 6000 kg·m/s / 800 kg
= 7.5 m/s
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A Caris travelling along astraigh levels red at 20 m/s against force of 3000M What Power forms its engine is needed?
Answer:
600kW
Explanation:
Power= Workdone/ time
= 1500/8*320
= 1500*40
= 60000J/s
= 600kW
Workdone= Fd
= 3000*1*1/16
= 1500/8
= 750/4
= 137. 5Nm
3000F/320
=150F/16
s=ut+1/2at^2
3000= 1/2at^2
6000= at^2
6000/a=t^2
F=ma
20m/t=ma
20/t= a
20m=Ft
20m=F(320)
m= 8F
F=ma
= 20/tm
20m/t= 20/tm
m= 1/m
m=1kg
6000/a= 400/a^2
16= 1/a
a= 1/16ms-2
t= 20/1/16
t= 320 s
(v-u)/t=a
v= 20ms-1
Glucose is a reactant in cellular respiration.
True or False?
Answer:
true
Explanation:
Answer:
True
Explanation:
Glucose is one of the reactants in cellular respiration, which is the process by which cells generate energy in the form of ATP (adenosine triphosphate). Glucose is broken down into simpler molecules in a series of metabolic reactions that occur in the presence of oxygen (aerobic respiration) or in the absence of oxygen (anaerobic respiration). The breakdown of glucose ultimately results in the release of energy that is used to produce ATP, which is the main source of energy for cellular activities.
Why is it important to assess your fitness level?
Assessing your fitness level is important because it help in tracking your progress and determine if you are making improvements. Regular assessments can help you identify areas where you may need to make adjustments to your fitness routine to achieve your goals. By assessing fitness level, you can identify areas where you may be weaker or less flexible. This information can help you design a fitness routine that addresses these areas and reduces our risk of injury.
Regular physical activity and exercise can improve overall health and reduce the risk of chronic diseases such as heart disease, diabetes, and obesity. By understanding your fitness level, you can design an exercise routine that helps you achieve optimal health and wellness.
From these measurements, compute the possible error in the volume.
Plugging in the values of the length and diameter, along with the error value, the error in the volume of the cylinder is approximately 1.27 cm³.
What is error in measurement?Error in measurement refers to the deviation or difference between the true or expected value and the measured value of a physical quantity. The presence of errors in measurement can affect the accuracy and precision of the results obtained.
To compute the possible error in the volume of the cylinder, we first need to calculate the volume of the cylinder using the measured values of its length and diameter:
V = πr²h
r = d/2 = 2.1/2 = 1.05 cm
V = π(1.05)²(8.9) = 31.79 cm³
Now, we need to determine the possible error in the volume, which can be calculated using the formula:
ΔV = V × √[(Δd/d)² + (Δh/h)²]
where Δd and Δh are the uncertainties in the diameter and length measurements, respectively. Substituting the given values, we get:
ΔV = 31.79 × √[(0.1/2.1)² + (0.1/8.9)²] = 1.27 cm³ (approx.)
Therefore, the possible error in the volume of the cylinder is approximately 1.27 cm³.
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A contractor is pushing a stove across a kitchen floor with a constant velocity of 18 cm/s [fwd]. The contractor is exerting a constant horizontal force of 85 N [fwd]. The force of gravity on the stove is 447 N [down].
Determine the normal force (FN ) and the force of friction (Ff ) acting on the stove.
Determine the total force applied by the floor (Ffloor) on the stove.
Answer:
Explanation:
Since the stove is moving with a constant velocity, we know that the net force on the stove is zero. Therefore, the force of friction acting on the stove must be equal in magnitude and opposite in direction to the horizontal force being applied by the contractor. We can use Newton's second law to solve for the normal force and force of friction:
ΣF = ma
where ΣF is the net force, m is the mass of the stove, and a is the acceleration of the stove (which is zero in this case).
First, we need to convert the velocity to m/s and the forces to Newtons (N):
18 cm/s = 0.18 m/s
85 N [fwd] - force applied by contractor
447 N [down] - force of gravity on the stove
Now we can solve for the normal force:
ΣFy = 0 (since the stove is not accelerating in the y-direction)
FN - 447 N = 0
FN = 447 N
Therefore, the normal force acting on the stove is 447 N.
Next, we can solve for the force of friction:
ΣFx = 0 (since the stove is moving at a constant velocity)
Ff - 85 N = 0
Ff = 85 N [bkwd]
Therefore, the force of friction acting on the stove is 85 N [bkwd].
Finally, we can solve for the total force applied by the floor:
ΣF = ma = 0 (since the stove is not accelerating)
Ffloor - 85 N - 447 N = 0
Ffloor = 532 N [up]
Therefore, the total force applied by the floor on the stove is 532 N [up].
A 208g sample of sodium-24 decays to 13.0g of sodium-24 within 60.0 hours. What is the half life of this radioactivity isotope?
Answer:
15 hours
Explanation:
formula: f(a) = a(0.5)^(T/t)
fill in known values: 13=208(0.5)^(60/t)
use natural log to isolate t: ln(13/208)=ln(0.5)(60/t)
solve for t: t=15
a boy throws a ball horizontally from shoulder height of 1.10m just before the ball touches down on the level ground it makes an angle of 30 degree with the ground. determine the initial velocity of the ball as it left the boys hand
The boy throws the ball horizontally. The initial velocity of the ball as it left the boy's hand was approximately 3.72 m/s.
What is initial velocity?Initial velocity, often represented as v0, is the velocity of an object at the beginning of a time interval or at the start of a motion.
Use the following kinematic equations to arrive at the answer:
Horizontal velocity (Vx) = Distance / Time
Vertical displacement (y) = V0y*t + (1/2)gt²
Vertical velocity (Vy) = V0y + g*t
Tan(theta) = Vy / Vx
where V0y is the initial vertical velocity, g is acceleration due to gravity (9.8 m/s²), and theta is the angle of inclination.
First, let's find the time it takes for the ball to hit the ground. We can use the vertical displacement equation and set y = 0:
0 = V0y*t + (1/2)gt²
Simplifying and solving for t, we get:
t = sqrt((2y) / g)
= sqrt((21.10 m) / 9.8 m/s²)
= 0.472 s
Now, we can use the horizontal velocity equation to find Vx. Since the ball was thrown horizontally, Vx is the same as the initial velocity (V0):
Vx = Distance / Time
= (horizontal distance travelled by ball) / t
We don't know the horizontal distance travelled by the ball, but we can find it using the vertical displacement equation. At the instant the ball hits the ground, its vertical displacement (y) is:
y = V0y*t + (1/2)gt²
= 0 + (1/2)gt²
= (1/2)*9.8 m/s² * (0.472 s)²
= 1.10 m
This means the ball travelled a total distance of:
distance = horizontal distance + vertical distance
= x + 1.10 m
where x is the horizontal distance travelled by the ball. We can find x using the angle of inclination and the vertical displacement:
Tan(theta) = Vy / Vx
Vy = V0y + g*t
Solving for V0y, we get:
V0y = Vy - g*t
Plugging in the numbers, we get:
V0y = Tan(theta) * Vx - g*t
= Tan(30 deg) * Vx - 9.8 m/s² * 0.472 s
= 0.577 * Vx - 4.62 m/s
Now, we can use the vertical displacement equation again to find x:
y = V0yt + (1/2)gt²
= (0.577Vx - 4.62 m/s) * 0.472 s + (1/2)*9.8 m/s² * (0.472 s)²
= 1.10 m
Simplifying and solving for Vx, we get:
Vx = (2y - 0.577V0t) / t
= (21.10 m - 0.577*(0.577*Vx - 4.62 m/s)*0.472 s) / 0.472 s
= 3.72 m/s
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A soccer player kicks a ball of mass 0.500 kg toward the goal.The ball hits the crossbar at a height of 2.6 m with a speed of 15.0m/s. Suppose the ball was at rest on the ground before it was kicked. Use g = 9.80 m/s.
Answer:
The speed of the ball just before it hits the crossbar is 7.22 m/s.
Explanation:
We can use the conservation of energy to solve this problem.
At the moment the player kicks the ball, the ball has only kinetic energy, since it was at rest on the ground before being kicked. When the ball hits the crossbar, it has both kinetic energy and potential energy, since it is at a height above the ground. We can set the initial kinetic energy equal to the sum of the final kinetic and potential energy:
(1/2)mv^2 = mgh
where:
m = mass of the ball (0.500 kg)
v = initial speed of the ball (15.0 m/s)
g = acceleration due to gravity (9.80 m/s^2)
h = height of the crossbar above the ground (2.6 m)
We want to solve for the speed of the ball just before it hits the crossbar, which we can do by rearranging the equation:
v = sqrt(2gh)
v = sqrt(29.802.6) = 7.22 m/s (rounded to two decimal places
PLEASE HELP, IM CONFUSED AND THIS IS A LATE ASSIGNMENT
Answer: 4 N backwards
Explanation:
A block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . determine the displacement of the velocity of the first block 1.2 s after the release of the blocks, assuming the first block doesn't run out of the room on the table and the second block doesn't land on the floor?
A) 23.5m/s
B) 12m/s
C) 33.7m/s
D) 6.7m/s
To solve this problem, we can use the principles of Newton's laws of motion and the conservation of energy.
At the moment of release, the second block will start to accelerate downwards due to gravity, and the first block will start to move to the right due to the tension in the rope. Since the surface is frictionless, there is no horizontal force acting on the first block once it starts moving.
Using the free-body diagrams for the two blocks, we can write the following equations of motion:
For the second block:
m2g - T = m2a
where g is the acceleration due to gravity, T is the tension in the rope, a is the acceleration of the second block, and m2 is the mass of the second block.
For the first block:
T = m1a
where m1 is the mass of the first block and a is its acceleration.
Since the two blocks are connected by a rope, they must have the same acceleration, so we can set the two equations for acceleration equal to each other:
m2g - T = m1a
T = m1a
m2g - m1a = T = m1a
Solving for a, we get:
a = (m2/m1 + m2)g
We can also use the conservation of energy to find the final velocity of the first block after 1.2 seconds. At the moment of release, the total mechanical energy of the system is given by:
E = m1gh
where h is the initial height of the second block. As the blocks move, the potential energy of the second block is converted into the kinetic energy of both blocks. At the end of the 1.2 seconds, all of the potential energy will be converted into kinetic energy, so we can write:
E = (1/2)m1v^2 + (1/2)m2v^2
where v is the final velocity of the first block.
Solving for v, we get:
v = sqrt(2gh(m1+m2)/m1)
Plugging in the given values, we get:
a = (2/5)g ≈ 3.92 m/s^2
v = sqrt(2gh(m1+m2)/m1) ≈ 2.36 m/s
Therefore, the displacement of the velocity of the first block 1.2 s after the release of the blocks is approximate:
vt + (1/2)at^2 = 2.361.2 + (1/2)3.92(1.2)^2 ≈ 5.52 m/s
So the answer is not given in the options.
an object is launched at a velocity of 40m/s in a direction making an angle of 50°upward with the horizontal
a)what is the maximum height reached by the object
b) what is the object total flight time between launch and touching the ground
c) what's the object horizontal range(maximum ×above ground)
Which has more kinetic energy:
a. A compact car going 70 MPH or a tractor trailer going 15 MPH?
b. An SUV going 30 MPH or a pickup truck going 30 MPH?
c. A school bus going 15 MPH, an SUV going 35 MPH, or a
compact car going 45 MPH?
Answer:
Explanation:
a. The compact car going 70 MPH has more kinetic energy than the tractor trailer going 15 MPH. Kinetic energy is proportional to the square of the velocity, so even though the tractor trailer may have more mass, the higher velocity of the compact car results in greater kinetic energy.
b. The SUV and the pickup truck have the same kinetic energy since they have the same mass and velocity.
c. The compact car going 45 MPH has the most kinetic energy since it has the highest velocity out of the three vehicles. The SUV going 35 MPH has less kinetic energy, and the school bus going 15 MPH has the least kinetic energy.
Why do you think the pylon in Figure 24 is designed the way it is, and not in the way shown in Figure 25?
They are specifically made tο be ideal fοr cοnducting live electrical lines because οf their electrical insulatiοn and mechanical tοughness. A structure called an electric pylοn οf hοt-rοlled steel bevels οr gusset plates.
What kinds οf patterns are used tο create electrical pylοns?Other materials, such as cοncrete and wοοd, may alsο be utilised in additiοn tο steel. Transmissiοn tοwers can be divided intο fοur main categοries: suspensiοn, terminal, tensiοn, οr transpοsitiοn.
Whο was the electrical pylοn's designer?This Central Electricity Bοard held a cοmpetitiοn in 1927, and the winning entry was chοsen by the classical designer Sir Reginald Blοοmfield. He settled οn an A-frame structure with latticewοrk that was οffered by the American cοmpany Milliken Brοthers and is still in use tοday.
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Complete question:
Question 14 of 16
When is work negative?
OA. When an object goes from high to low potential energy
O B. When an object goes from low to high potential energy
O C. When an object slows down
O D. When an object speeds up
Work is negative when A. When an object goes from high to low potential energy.
When can work be said to be negative ?Work is defined as the transfer of energy that occurs when a force is applied to an object, causing it to move a certain distance. When the force and displacement are in the same direction, the work is positive.
When an object moves from a higher to a lower potential energy state, it loses potential energy. In this case, the negative work done by gravity is equal to the loss of potential energy of the ball.
When an object slows down, the force acting on it is in the opposite direction to its velocity, so work is done in the opposite direction to the displacement, and the work is negative.
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help me
Write your answer on the lines below.
4. Are the light waves reflecting off a red stop sign longer or shorter than the waves reflecting off a violet-colored jacket? Explain how you know.
The light waves reflecting off a red stop sign would be longer than the light waves reflecting off a violet-colored jacket. This is because red light has a longer wavelength than violet light.
Light waves and reflectionLight waves, like all waves, are characterized by their wavelength. The wavelength of a wave determines its color, with shorter wavelengths appearing as blue and violet, and longer wavelengths appearing as red and orange.
Because the wavelength of red light is longer than the wavelength of violet light, the light waves reflecting off the red stop sign would be longer than the light waves reflecting off the violet-colored jacket.
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Owen hits a baseball with a velocity of 55 m/s. The ballpark fence is 120 m away.
Does the ball reach the fence if it leaves the bat traveling upward at an angle of 30°
to the horizontal?
Answer:
Explanation:
We can solve this problem using kinematic equations. We know that the initial velocity of the ball is 55 m/s at an angle of 30° to the horizontal. We can break this velocity into its horizontal and vertical components:
vx = v0 cos θ = 55 cos 30° = 47.6 m/s
vy = v0 sin θ = 55 sin 30° = 27.5 m/s
We can now use the vertical motion equation to find the time it takes for the ball to reach its maximum height:
Δy = vy t + 0.5 a t^2
At the maximum height, the vertical velocity of the ball is 0, so we have:
0 = vy + a t_max
Solving for t_max, we get:
t_max = -vy / a = -27.5 / (-9.8) = 2.81 s
The ball will take twice this time to reach the fence, since it needs to come back down to the ground:
t_total = 2 t_max = 5.62 s
The horizontal distance the ball travels during this time is:
Δx = vx t_total = 47.6 × 5.62 = 267.7 m
Since this distance is greater than the distance to the fence (120 m), the ball will reach the fence if it leaves the bat traveling upward at an angle of 30° to the horizontal.
A uniformly charged insulating sphere with radius r and charge +Q
lies at the center of a thin-walled hollow cylinder with radius R>r
and length L>2r. The cylinder is non-conducting and carries no net charge.
1:Determine the outward electric flux through the rounded "side" of the cylinder, excluding the circular end caps. (Hint: Choose a cylindrical coordinate system with the axis of the cylinder as its z -axis and the center of the charged sphere as its origin. Note that an area element on the cylinder has magnitude dA=2πRdz
2:Determine the electric flux upward through the circular cap at the top of the cylinder.
3:Determine the electric flux downward through the circular cap at the bottom of the cylinder.
4:Add the results from parts A - C to determine the outward electric flux through the closed cylinder.
5:What result is expected according to Gauss's law?
Note:Express your answers in terms of electric constant ϵ0
and some or all of the variables r, R , L , Q .
According to Gauss' equation, the total flux of an electric field in a confined surface is directly proportional to the charge enclosed.
State Gauss’s law.1)To determine the outward electric flux through the rounded "side" of the cylinder, we can use Gauss's law. We choose a cylindrical Gaussian surface with radius r and length L, centered at the origin (where the charged sphere is located). The electric field due to the sphere is spherically symmetric, so the electric field lines are parallel to the cylinder's axis and perpendicular to its sides.
E = (1/4πϵ0) (Q/r^2)
where r is the distance from the origin (center of the sphere) to the point on the Gaussian surface.
The area element of the Gaussian surface is dA = 2πRdz, where dz is an element of length along the cylinder's axis. The electric flux through the top and bottom surfaces of the Gaussian surface is then given by:
Φ = ∫E⋅dA = E ∫dA = E(2πR)L
Substituting the expression for the electric field, we have:
Φ = (Q/2ϵ0r^2)(2πRL)
Therefore, the outward electric flux through the rounded "side" of the cylinder is:
Φ = (Q/ϵ0)(R/Lr^2)
2)To determine the electric flux upward through the circular cap at the top of the cylinder, we use a flat Gaussian surface with radius R and height r, centered at the top of the cylinder. The electric field due to the charged sphere is perpendicular to the Gaussian surface, so the electric flux through the top cap is simply the flux through the flat Gaussian surface. The electric field at any point on the Gaussian surface is given by Coulomb's law as:
E = (1/4πϵ0) (Q/R^2)
The area element of the Gaussian surface is dA = πR^2, so the electric flux through the top cap is given by:
Φ = ∫E⋅dA = E ∫dA = EπR^2
Substituting the expression for the electric field, we have:
Φ = (Q/ϵ0)(R/r^2)
3)To determine the electric flux downward through the circular cap at the bottom of the cylinder, we use a similar flat Gaussian surface with radius R and height r, centered at the bottom of the cylinder. The electric flux through the bottom cap is also given by:
Φ = (Q/ϵ0)(R/r^2)
4)Adding the results from parts 1-3, we have the total outward electric flux through the closed cylinder as:
Φ_total = Φ_side + Φ_top + Φ_bottom
= (Q/ϵ0)(R/Lr^2) + 2(Q/ϵ0)(R/r^2)
Simplifying this expression, we have:
Φ_total = (Q/ϵ0) [(2R/r^2) + (R/Lr^2)]
5)According to Gauss's law, the total outward electric flux through a closed surface is proportional to the total charge enclosed within that surface. In this case, the closed surface is the cylindrical Gaussian surface with radius r and length L, centered at the origin (where the charged sphere is located). The charge enclosed within this surface is simply the charge of the sphere, which is +Q. Therefore, we expect the total outward electric flux through the closed cylinder to be:
Φ_total = Q/
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State each of Newton's Laws of Motion and explain how each can be observed during the flight of a space craft, from liftoff until the craft enters space.
newton 3th law of motion and newton's law of universal gravitation
Answer: 1. Newton's First Law of Motion (Law of Inertia): An object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by an external force.
During liftoff, the spacecraft is initially at rest. However, the rocket engines generate a force that propels the spacecraft forward and overcomes its initial state of rest. Once the spacecraft is in motion, it will continue to move forward with a constant velocity unless acted upon by other external forces, such as air resistance or gravity.
2. Newton's Second Law of Motion: The acceleration of an object is directly proportional to the force applied to it, and inversely proportional to its mass.
As the rocket engines burn fuel, they generate a force that propels the spacecraft forward. The acceleration of the spacecraft is directly proportional to the force generated by the engines, and inversely proportional to the mass of the spacecraft. As fuel is consumed and the spacecraft becomes lighter, its acceleration will increase, allowing it to reach escape velocity and enter space.
3. Newton's Third Law of Motion: For every action, there is an equal and opposite reaction.
During liftoff, the rocket engines generate a powerful force that propels the spacecraft forward. However, the engines also generate an equal and opposite reaction force, pushing back against the rocket and causing it to shake and vibrate. This force is also responsible for the loud noise and exhaust plumes that are visible during liftoff.
These are the three laws of motion developed by Sir Isaac Newton, and they explain how objects move and interact with one another. They can be observed in the launch and flight of a spacecraft, from the initial state of rest to the forces that drive it forward, to the equal and opposite forces that shake the rocket during liftoff.
Express your answer with the appropriate units. A 60.0 kg box hangs from a rope. What is the tension in the rope if:
A. The box is at rest?
B. The box moves up a steady 4.80 m/s ?
C. The box has vy = 5.00 m/s and is speeding up at 5.40 m/s^2 ? The y axis points upward.
D. The box has vy = 5.00 m/s and is slowing down at 5.40 m/s^2 ?
When the box is at rest, the tension in the rope is equal to the weight of the box. The tension in the rope is 588 N.
What is tension in physics?In physics, tension refers to the pulling force that is transmitted through a string, cable, rope, etc when it is pulled tight by forces acting at both ends. Tension is a vector quantity, and measured in units of newtons (N) or pounds (lbs).
A. When the box is at rest, the tension in the rope is equal to the weight of the box, which is given by:
Tension = Weight of the box = mg = (60.0 kg)*(9.81 m/s²) = 588 N
Thus, the tension = 588 N.
B. When the box moves up at a steady 4.80 m/s, the tension in the rope is equal to the force required to lift the box against gravity, which is given by:
Tension = Weight of the box + Force to lift the box = mg + ma = (60.0 kg)*(9.81 m/s²) + (60.0 kg)*(4.80 m/s²) = 1,167.6 N
Therefore, the tension in the rope is 1,167.6 N.
C. When the box has velocity along the y-axis = 5.00 m/s and is speeding up at 5.40 m/s², the tension in the rope is given by the equation:
Tension = Weight of the box + Force to accelerate the box = mg + ma = (60.0 kg)*(9.81 m/s²) + (60.0 kg)*(5.40 m/s²) = 1,199.4 N
Therefore, the tension in the rope is 1,199.4 N.
D. When the box has velocity along y-axis = 5.00 m/s and is slowing down at 5.40 m/s², the tension in the rope is given by the equation:
Tension = Weight of the box - Force to decelerate the box = mg - ma = (60.0 kg)(9.81 m/s²) - (60.0 kg)(5.40 m/s²) = 981.6 N
Therefore, the tension in the rope is 981.6 N.
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A firefighting crew uses a water cannon that shoots water at 29.0 m/s
at a fixed angle of 53.0 ∘
above the horizontal. The firefighters want to direct the water at a blaze that is 10.0 m
above ground level. How far from the building should they position their cannon? There are two possibilities ( d1
); can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)
Express your answer in meters.
d1 and d2
The second possibility is that the firefighters can position their cannon anywhere within a range of 60.2 m from the building to hit the blaze at 10.0 m above ground level.
What is Horizontal Velocity?
Horizontal velocity is the component of velocity in the horizontal direction, perpendicular to the vertical direction. It describes the speed and direction of motion of an object in the horizontal plane. In the absence of external forces, the horizontal velocity of an object moving through the air remains constant, as there is no force acting on the object in the horizontal direction.
Let's first find the time it takes for the water to reach the maximum height:
The vertical component of the initial velocity is:
vy = v * sin(θ) = 29.0 m/s * sin(53.0°) = 22.7 m/s
Using the kinematic equation:
where Δy = 10.0 m, vyi = 22.7 m/s, and a = -9.81 m/[tex]s^{2}[/tex] (the negative sign indicates that acceleration is in the opposite direction to the initial velocity).
We get:
10.0 m = 22.7 m/s * t - 1/2 * 9.81 m/[tex]s^{2}[/tex] * [tex]t^{2}[/tex]
Simplifying and solving for t, we get:
t = 1.61 s
Now, let's find the horizontal displacement of the water:
The horizontal component of the initial velocity is:
vx = v * cos(θ) = 29.0 m/s * cos(53.0°) = 18.7 m/s
Using the equation:
Δx = vx * t
where Δx is the horizontal displacement, vx is the horizontal component of the initial velocity, and t is the time we found above.
We get:
Δx = 18.7 m/s * 1.61 s = 30.1 m
So the firefighters should position their cannon 30.1 m away from the building.
To find the second possibility, we need to find the range of the water cannon, which is the horizontal distance traveled by the water before it hits the ground. The range can be calculated using the formula:
where R is the range, v is the initial velocity, θ is the angle above the horizontal, and g is the acceleration due to gravity.
Plugging in the values we get:
R = [tex]29.0^{2}[/tex] * sin(2 * 53.0°) / (2 * 9.81) = 60.2 m
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What is the electric potential energy of the group of charges in the figure? (Figure 1)
that the relative placements of the charges as well as their multiples affect a set of ions' potential energy. When the specific charge have the same sign or have equal signs, the energy is positive. Or else, it is negative.
How is potential energy calculated?The force acting just on two objects affects the potential energy formula. The formula for gravitational force is P.E. (= mgh, where g seems to be the acceleration caused by gravity (9.8 m/s2 at the earth's surface) while h represents the elevation in metres.
What is a system with two charges' potential energy?As a result, the system's potential energy equals the sum of a work that was done to set up the entire system of two counts. The potential energy that exists in the combination of two charges in such an external field can be stated as follows: q1V(r1) = q2V(r2) + (q1q2/4or12).
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A crate of mass
m = 26 kg
rides on the bed of a truck attached by a cord to the back of the cab as in the figure below. The cord can withstand a maximum tension of 69 N before breaking. Neglecting friction between the crate and truck bed, find the maximum acceleration the truck can have before the cord breaks. (Enter the magnitude of the maximum acceleration in the forward direction.)
m/s2
Answer:
Explanation:
The maximum tension the cord can withstand is 69 N, so we know that the tension in the cord cannot exceed this value. The tension in the cord is related to the acceleration of the truck through Newton's second law:
ΣF = ma
where ΣF is the net force on the crate, m is the mass of the crate, and a is the acceleration of the truck.
In this case, the only force acting on the crate in the horizontal direction is the tension in the cord. Therefore, we can write:
ΣF = T = ma
where T is the tension in the cord.
We can solve this equation for the acceleration:
a = T/m
We know that the tension cannot exceed 69 N, so the maximum acceleration the truck can have before the cord breaks is:
a = 69 N / 26 kg
a ≈ 2.65 m/s^2
Therefore, the maximum acceleration the truck can have before the cord breaks is 2.65 m/s^2.
The units of voltage are
Oohms, amps, volts
Oamps, volts, ohms
Ovolts, ohms, amps
Ovolts, amps, ohms
of current are
and of resistance are
Answer:24 volts ÷ 4 amps = 6 ohms
Explanation:
We know that the Ohm's law has given the relationship between the current, voltage and the resistance of the wire. Mathematically, it can be written as :
Where
I is the current
R is the resistance
If current flowing is 4 amps and voltage is 24 volts. The formula to find the resistance will be :
R = 6 Ohms
Hence, the correct option is (d) " 24 volts ÷ 4 amps = 6 ohms ".
What is the momentum of a 2.3 kg ball rolling at 6 m/s?
Show your work
Answer:
13.8 (kgm)/s
Explanation:
p(momentum) = m (mass) * v (velocity)
p= 2.3 * 6
p = 13.8