A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60
what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.
Answer:
Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].
If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].
Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.
Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)
By Newton's second law of motion, the acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].
In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].
Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.
A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?
Hi there!
We can begin by calculating the acceleration of the block and the wheel using the following equation:
d = vit + 1/2at², where initial velocity = 0 m/s
d = 1/2at²
2d/t² = a
2(1.5)/2² = 0.75 m/s²
Now, we can do a summation of torques:
∑τ = rT
Rewrite using Newton's 2nd Law for rotation:
Iα = rT
Convert α to a using the relationship α = a/r:
I(a/r) = rT
Ia = r²T
I = r²T/a
Plug in the values:
I = (0.40²)(20)/(0.75) = 4.267 kgm²
2) A rolling disk, mass m and radius R, approaches a step of height R/2 with velocity v. (i) Taking the corner of the step as the pivot point, what is the initial angular momentum of the disk
The rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
Using the law of conservation of energy, the initial mechanical energy E of the disk equals its final mechanical energy E' as it climbs the step.
So, E = E'
1/2Iω + 1/2mv² + mgh = 1/2Iω' + 1/2mv'² + mgh'
where I = rotational inertia of disk = 1/2mR² where m = mass of disk and R = radius of disk, ω = initial angular speed of disk, v = initial velocity of disk, h = initial height of disk = 0 m, ω' = final angular speed of disk = 0 rad/s (assumung it stops at the top of the step), v' = final velocity of disk = 0 m/s (assumung it stops at the top of the step), and h' = final height of disk = R/2.
Substituting the values of the variables into the equation, we have
1/2Iω² + 1/2mv² + mgh = 1/2Iω'² + 1/2mv'² + mgh'
1/2(1/2mR² )ω² + 1/2mv² + mg(0) = 1/2I(0)² + 1/2m(0)² + mgR/2
mR²ω²/4 + 1/2mv² + 0 = 0 + 0 + mgR/2
mR²ω²/4 + 1/2mv² = mgR/2
R²ω²/4 = gR/2 + 1/2v²
R²ω²/4 = (gR + v²)/2
ω² = 2(gR + v²)/R²
ω² = √[2(gR + v²)/R²]
ω = √[2(gR + v²)]/R
Since angular momentum L = Iω, the rolling disk's initial angular momentum is
L = 1/2mR² ×√[2(gR + v²)]/R
L = mR√[2(gR + v²)]/2
the rolling disk's initial angular momentum is mR√[2(gR + v²)]/2
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A 1300 watt hair blow dyer is designed to operate on 120 Volts. How much current does the dryer require
Answer:
10.83 Amperes
Explanation:
if A ⇒ current
W = VA
1300 = 120 x A
1300 / 120 = A
10.83 = A
A friend has suggested that you go swimming in a pool having water of temperature 350 K. What would this temperature be on the Fahrenheit scale?
109°F
123°F
170°F
202°F
This temperature would be 170° F on the Fahrenheit scale. Hence, option (C) is correct.
What is temperature?The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points and thermometric substances.
The most popular scales are the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the Fahrenheit scale (°F), and the Kelvin scale (K), with the latter being mostly used for scientific purposes.
the relation between Kelvin scale and Fahrenheit scale is given by:
(F - 32)/180 = (K - 273)/100
F - 32 = (350 - 273)(9/5)
F = 32 + (350 - 273)(9/5)
F = 170
Hence, this temperature would be 170° F on the Fahrenheit scale.
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please answer this as fast as you can i need it
Answer:
it says pdf only i dont knowwhat u want me to do
g What is the CD's moment of inertia for rotation about a perpendicular axis through the edge of the disk
Answer:
Explanation:
A CD has an OD of 120 mm and an ID of 15 mm and has a mass between 14 and 33 grams. Let's call it m
Lets call the outer and inner radii R and r respectively
Find the moment of inertia about a line perpendicular to the surface of the disc through its center. We can integrate or look up the result from standard tables
I = ½m(R² + r²)
then use the parallel axis theorem to shift the position of the axis
I = ½m(R² + r²) + md²
where d is the distance of the shift. In this case d = R
I = ½m(R² + r²) + mR²
I = m(1.5R² + 0.5r²)
If we select a mass of say 20 grams
I = 0.020(1.5(0.060²) + 0.5(0.0075²))
I = 0.0001085625 kg•m²
A 6.5 N ball is thrown with an initial velocity of 20 m/s at a 35° angle from a height of 1.5 m, what is the velocity if it is caught at 1.5 m?
Answer:
20 m/s at -35°
Explanation:
Ignoring air resistance, the initial vertical velocity will be reversed and the initial horizontal velocity will remain constant.
Why is it important for the community to take action
Answer:
Why is community action important? ... Involving communities in the design and delivery of services can help to achieve a number of objectives, including: Building community and social capacity – helping the community to share knowledge, skills and ideas. Community resilience – helping the community to support itself
Convert :
36°C = ... °F
373 K = ... °C
Question easy
Answer:
36 C= 96.8 F
373 K= 99.85
Explanation:
C to F: (36 x 1.8) + 32
= 64.8 +32
= 96.8 F
K to C: C= K- 273.15
C= 373-273.15
C= 99.85
____
= 36°C
=( 36 × 9/5 ) + 32
=(36 ÷ 5 × 9) + 32
=(7,2 × 9) + 32
= 64,8 + 32
= 96,8°F______
______
= 373 K
= 373 - 273
= 100°C[tex] \boxed { \sf semoga \: membantu \: :v }[/tex]
in rotational motion the angular frequency of all the particles is
Answer:
L=mvr
α=0
v=ωr
L=mωr
2
2
1
mv
2
=k
L=Iω
k=
2
1
Iω
2
ω
1
=2ω
k
1
=1/2k
∴I
1
=
8
1
I
L
1
=
8
1
I×2ω
=
4
1
Iω
=L/4
Physical motion that occurs when an item rotates or spins on an axis is known as “rotatory motion,” sometimes known as “rotational motion” or “circular motion.”
What angular frequency of particles in rotational motion?Angular displacement per unit of time is measured by angular frequency, also referred to as radial or circular frequency (). Therefore, it has degrees (or radians) per second as its units. 1 Hz = 6.28 rad/sec, thus. 1 radian equals 57.3° because 2 radians equal 360°.
The frequency of rotation is the rate at which an item rotates around an axis, sometimes referred to as rotational speed or rate of rotation.
However, the linear velocity also depends on how far the particles are from the axis of rotation. Because every particle distances are different, so too will the linear velocities for each particle.
Therefore, All the particles in a rigid body's rotational motion move in the same direction in a given interval. Therefore, all the particles' angular velocities will be the same.
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I NEED THE ANSWER ASAPP
Answer:
Explanation:
a) The spring force will equal the weight.
b) If up is positive
kx - mg = 0
mg = kx kx = 25 N
c) m = kx/g = 25/10 = 2.5 kg
A rock is thrown off a cliff with a speed of 5 m/s downward. How far will it fall after 7 seconds have elapsed?
Free-fall Acceleration is -10 m/s^2
I also need the Formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
if the throw point is origin and UP the positive direction and ignoring air resistance.
s = 0 + (-5)(7) + ½(-10)(7²)
s = 0 - 35 - 245
s = - 280 m
Anita Knapp needs to get hay to cows in a frozen field using an airplane flying
80.0 m/s, at a height of 300,m. If at the last minute, how far from the cow would
she have to release the hay in order to hit the cow?*
756 m
626m
700m
575 m
Other:
Answer:
626m
Explanation:
A metal wire of length 1.2 m and cross-sectional area 2.0 x 10 raised to power-7 m 2 is stretched by a force of 50 N. assuming the force constant of the metal is 6000 Nm-1. Calculate the tensile stress
L=1m
=2mm²=(2/1000²)=2(10^-6)m
y=4x10¹¹N/m²
∆l=2mm=2/10.00=0.002m
=(4x10¹¹x2x10^-6x2x10^-3)÷1m
=16x10¹¹-⁶-³
=16x10¹¹-⁹
=16x10²
=1600N
where a=cross sectional area=2x10^-6m
C=2mm= 2x10^-3m
L=1m
hope it helps
Which statement describes electromagnetic waves with wavelengths grater than 700 nanometers
Answer:
They take the form of heat, I think thats it but I cant see if you put down any answers
Explanation:
Answer:a
Explanation:
they form hear
This is not a question
do your work in class you would know kids
a boat's engine can give it a velocity of 25m/s. if the boat heads east across a river which of the following due south with a velocity of 8.5m/s; what is the resultant velocity of the boat? (remember you must find both a magnitude and direction!)
Answer:
Explanation:
v = √(25² + 8.5²) = 26.40549... = 26 m/s
θ = arctan(8.5/25) = 18.77803... = 19° S of E
The figure shown above is the circuit diagram for a simple dc power supply. Identify the type of rectifier circuit represented in the figure and explain the operation of the circuit with reference to the function of each component within the circuit.
Answer:
D1 FG 12 15×AG+5T×G7+3F
2. Which of the following contributions did Louie De Broglie do for electronic structure of matter? A. determined the speed of electron of hydrogen atom B. proposed a theory that electrons showed characteristics similar to light C. provided mathematical operation for the characteristics of light D. recorded the movement of proton in the nucleus of an atom
❤️
Answer:
In 1924 Louis de Broglie introduced the idea that particles, such as electrons, could be described not only as particles but also as waves. This was substantiated by the way streams of electrons were reflected against crystals and spread through thin metal foils.
Explanation:
I know I probably didn't answer your question, I just used all of my knowledge that I learned about Louie De Broglie. Hope it helps!
A 100 N crate is being pulled at a constant velocity by a rope a 30 degrees to the horizontalas depicted in the diagramFind the force of friction Show your work and explain your reasoning in two to sentences
Answer:
Explanation:
As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.
Ff = Fcosθ
if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.
μN = Fcosθ
μ(mg - Fsinθ) = Fcosθ
μmg = Fcosθ + μFsinθ
100μ = F(cos30 + μsin30)
F = 100μ / (cos30 + ½μ)
Ff = 100μcos30 / (cos30 + ½μ)
A 100 N create is being pulled at a constant velocity by a rope a 30 degrees to the horizontal as depicted in the diagram given in question the force of friction Ff = 100μcos30 / (cos30 + ½μ).
What is force?
A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.
Ff = Fcosθ
if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.
μN = Fcosθ
μ(mg - Fsinθ) = Fcosθ
μmg = Fcosθ + μFsinθ
100μ = F(cos30 + μsin30)
F = 100μ / (cos30 + ½μ)
Ff = 100μcos30 / (cos30 + ½μ)
the force of friction Ff, is 100μcos30 / (cos30 + ½μ).
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Need help with dot product
[tex]\textbf{A}\cdot\textbf{B} = 11.5[/tex]
Explanation:
The dot product between two vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] is defined as
[tex]\textbf{A}\cdot\textbf{B} = AB\cos{\theta}[/tex]
where A and B are the magnitudes of the vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B},[/tex] respectively and [tex]\theta[/tex] is the angle between the two. Since A = 3, B = 5 and [tex]\theta = 40°,[/tex] the dot product [tex]\textbf{A}\cdot\textbf{B}[/tex] is
[tex]\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5[/tex]
The center of mass of a 1600 kg car is midway between the wheels and 0.7 m above the ground. The wheels are 2.6 m apart. (a) What is the minimum acceleration A of the car so that the front wheels just begin to lift off the ground
Answer:
Explanation:
I guess we are ASSUMING that this is a rear wheel drive car as a front wheel drive car will never get the front wheel normal force to zero
If we consider it as a statics problem and choose our moment center carefully...say 0.7 m above the rear wheel to ground contact point.
Call the traction force at the rear wheels F
The normal force on the front wheels will be zero, so no moment generated by the front wheels.
Summing moments about our chosen point to zero
1600(9.8)[2.6 / 2] - F[0.7] = 0
F = 291,200
this force will create an acceleration of
a = F/m
a = 291200/1600
a = 182 m/s²
which is about 18.6 times gravity acceleration
Seven friends equally split a restaurant bill that
comes to $93.17. How much does each person pay?
Answer:
$13.31
Explanation:
We know that the bill comes to $93.17 and that 7 people will split the bill equally
We can just use the equation
bill = $93.17/7
bill = $13.31
Two trucks leave at different times (from the same place) headed for the same city. Both trucks arrive at the same time. Based on this information, which of the following sentences is true? Select one:
a. The trucks travelled the same distance in the same amount of time.
b. The trucks were traveling at the same average speed.
c. The trucks travelled different distances.
d. The truck that left later was travelling faster.
Answer:
d. The truck that left later was travelling faster
Explanation:
Both trucks travelled from the same place to the same place, meaning they both travelled the same distance;
They both arrive at the same time, but the second truck left later so it took less time to travel the distance than the first truck;
The only variable that can account for this difference is speed;
The one that left later, therefore, must have been going faster.
An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
Answer:
The distance traveled before takeoff is 1720 m
Explanation:
Given:a = + 3.2 m/s²
t = 32.8 s
Vᵢ = 0 m/s
To Find:d = ?
Now,
d = Vᵢ × t + 0.5 × a × t²
d = (0 m/s) × (32.8 s) + 0.5 × (3.20 m/s²) × (32.8 s)²
d = 1720 m
Thus, The distance traveled before takeoff is 1720 m
-TheUnknownScientist 72
Two blocks are set in a pully system as shown in fig below. Block A sits on the frictionless table while block B hags freely. The pully is light and frictionless towards the light string that runs over it. If the Block A has mass of 3.4 kg and Block has 3.5 kg, what would be the magnitude of the acceleration (in ms-2) of the blocks? [g = 9.8 ms=2]
Answer:
Explanation:
F = ma
a = F/m
a = mBg / (mB + mA)
a = 3.5(9.8)/(3.5 + 3.4)
a = 4.971014...
a = 5.0 m/s²
If you want to use individual Free Body Diagrams
mass A will have downward weight and upward normal forces equal at mAg
and a horizontal force of string tension T
F = ma
T = mAa
mass B will have a downward force of mBg and an upward force of T
mBg - T = mBa
substitute for T
mBg - mAa = mBa
mBg = a(mB + mA)
a = mBg / (mB + mA) which is identical to the above answer.
a Answer the following questions 1. On a cold wintery day, you burn firewood to keep yourself warm. The firewood undergoes a change in state. a. Name the change in state of matter that you see
Answer:
heating prosedure takes place the opposite of condensation
A tennis player strikes the tennis ball with an initial velocity of 44.7 m/s horizontally. The ball is initially 1.28 m above the ground and 12.9 m from the 0.914 m tall net. Does the tennis ball make it over the net?
Hi there!
We can begin by finding the total time taken for the ball to reach the net using the equation:
dₓ = vₓt
12.9 = 44.7t
12.9/44.7 = t = 0.289 s
Now, we can use the following equation to solve for displacement in the Y direction:
d = y₀ + vit + 1/2at²
There is no initial vertical velocity, so:
d = y₀ + 1/2at²
Plug in known values:
d = 1.28 + 1/2(-9.8)(0.289²)
d = 0.87m
Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.
What type of equilibrium maintains body position during sudden motion?
dynamic
rotational
static
balanced
I think static is the correct answer