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Diano4ka-milaya [45]

3 years ago

5

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un

iform acceleration of 2.75 m/s 2 . How much time passes before the speeder is overtaken by the police car? Answer in units of s.

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

$x=vt=23.3\frac{m}{s}t$

and the police car distance:

$x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}$

Since they both travel the same distance x, we can equal both formulas and solve for t:

$0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\$

Two solutions exist to the equation; the first one being $t=0$

The second solution will be:

$0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s$

This result allows us to confirm that the police car will take 32s to catch up to the speeder

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Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They co

balu736 [363]

The combined momentum of the passengers is 5000 kgm/s.

<h3>Combined momentum of the passenger</h3>

The combined momentum of the passengers is calculated as follows;

P = mv1 + mv2

where;

m is mass of the passengers
v1 is velocity of the first passenger
v2 is velocity of the second passenger

P = m(v1 + v2)

P = 5000(-1 + 2)

P = 5000 kgm/s

Thus, the combined momentum of the passengers is 5000 kgm/s.

Learn more about momentum here: brainly.com/question/7538238

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5 0

2 years ago

PLEASE ANSWER ASAP

Sedbober [7]

Answer:

B.

Explanation:

The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.

3 0

3 years ago

Turning a magnet very quickly would be BEST used to create A) radiation. Reactivate B) light waves. Reactivate C) an electric cu

Softa [21]

<h2>Answer:</h2>

<u>Turning a magnet very quickly would be BEST used to create an electric current</u>

<h2>Explanation:</h2>

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3 0

3 years ago

Read 2 more answers

Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

6 0

3 years ago

A 300-kg piano being held by a crane is accidentally dropped from a height of 15 meters. a. What is the speed of the piano just

FinnZ [79.3K]

Answer:

a) 17.16m/s

b) 44,145J

c) Sound the piano makes when hitting the ground, vibration of the ground, heat.

d) i) It's smaller due to the energy dissipated by the friction between air and the parachute.

ii) It stays the same, the only difference is that the dissipated energy is distributed between air resistance and the kinetic energy dissipated by the ground whent he piano hits it.

Explanation:

a)

In order to solve this problem we must start by doing a drawing of the situation, which will help us visualize the problem better. (See attached picture).

So, in this problem we can ignore air resistance so we can say that the energy is conserved, this is the total initial energy is the same as the total final energy, so we get that:

$U_{0}+K_{0}=U_{f}+K_{f}$

When the piano is released it has an initial speed of zero, so the initial kinetic energy is zero. When the piano hits the ground it will have a height of 0m, so the final potential energy is zero as well. This will simplify our equation:

$U_{0}=K_{f}$

We know that potential energy is given by the formula:

U=mgh

and kinetic energy is given by the formula:

$K=\frac{1}{2}mv^{2}$

which can be substituted in our equation:

$mgh=\frac{1}{2}mv^{2}$

we can divide both sides of the equation into the mass of the piano, so we get:

$gh=\frac{1}{2}v^{2}$

which can be solved for the final velocity which yields:

$v=\sqrt{2gh}$

we can now substitute the data provided by the problem so we get:

$v=\sqrt{2(9.81m/s^{2})(15m)}$

which yields:

v=17.16m/s

b)

Since energy is conserved, this means that the total dissipated energy will be the same as the potential energy, so we get that:

E=mgh

so

$E=(300kg)(9.81m/s^{2})(15m)$

which yields:

E=44,145J

c)

When the piano hits the ground, the kinetic energy it had will be transformed to other types of energy, mostly vibration and heat. The vibration will turn to sound due to the movement of air created by the piano itself and the ground. And heat is created by the friction between the molecules created by the vibrations and the collition itself. So some of the indicators of this release of energy could be:

-Sound

-Vibration

-Heat.

d)

i) The amount of inetic energy dissipated would decrease due to the friction between air and the parachute. Since air is resisting the movement of the piano, this will translate into a loss of energy, if we did an energy balance we would get that:

$U_{0}=K_{f}+E_{p}$

The total amount of energy is conserved but it will be distributed between the energy lost due to air resistance and the kinetic energy the piano has at the time it hits the ground.

ii) So the total amount of energy dissipated remains the same, the only difference is that it will be distributed between air resistance and the kinetic energy of the piano.

3 0

3 years ago