Ask question

Ask question

All categories

3 years ago

11

During a very quick stop, a car decelerates at 28.4 rad/s?. Assume the tires initially rotated in the positive direction and rad

ius of tires is 0.32m a) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 93.0 rad/s? b) How long does the car take to stop completely? c) What was the car's initial velocity (in m/s)? d) How far the car moves forward before it stops?

1 answer:

Damm [24]3 years ago

8 0

Answer:

a) 24

b) 3.3 sec

c) 29.8 m/s

d) 48.85 m

Explanation:

a)

α = angular acceleration = - 28.4 rad/s²

r = radius of the tire = 0.32 m

w₀ = initial angular velocity = 93 rad/s

w = final angular velocity = 0 rad/s

θ = angular displacement

Using the equation

w² = w₀² + 2αθ

0² = 93² + 2 (- 28.4) θ

θ = 152.3 rad

n = number of revolutions

Number of revolutions are given as

$n = \frac{\theta }{2\pi }$

$n = \frac{152.3 }{2(3.14) }$

$n = 24$

b)

t = time taken to stop

using the equation

w = w₀ + αt

0 = 93 + (- 28.4) t

t = 3.3 sec

c)

v₀ = initial velocity of the car

initial velocity of the car is given as

v₀ = r w₀ = (0.32) (93) = 29.8 m/s

d)

v = final velocity = 0 m/s

a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²

d = distance traveled by car before stopping

Using the equation

v² = v₀² + 2 a d

0² = 29.8² + 2 (- 9.09) d

d = 48.85 m

You might be interested in

Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of

STatiana [176]

Answer:

$2.27\cdot 10^{49}$

Explanation:

The gravitational force between the proton and the electron is given by

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p$ is the proton mass

$m_e$ is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N$

The electrical force between the proton and the electron is given by

$F_E=k\frac{q_p q_e}{r^2}$

where

k is the Coulomb constant

$q_p = q_e = q$ is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N$

So, the ratio of the electrical force to the gravitational force is

$\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}$

So, we see that the electrical force is much larger than the gravitational force.

5 0

3 years ago

Once potential energy is lost on a roller coaster it cannot be gained all the way back true or false

nignag [31]

Answer:

true

Explanation:

because the roller coaster can't work without energy

4 0

3 years ago

Helppppp pleaseee :(

Elza [17]

1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum

7 0

3 years ago

A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,

Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. <span>

An initial condition has been supplied by the problem:

s=1.80m when t=0.245s

<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2)

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation):

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2

v=8.549m/s

<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t)

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like

0=(v)+(a)(t)

Substituting and solving for t:

0=(8.549m/s)+(-9.81m/s^2)(t)

t=0.8714s

<span>Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m<span>

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window:

3.725m – 1.8m=1.925m</span>

Answer:

<span>1.925m</span>

3 0

3 years ago

Which is a characteristic of projectile motion?A force can affect the _____ of an object.

Olin [163]

Answer:

mass of an object

Explanation:

because mass of an object changes it’s force

5 0

2 years ago