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astraxan [27]
3 years ago
10

Car B has the same mass as Car A but is going twice as fast. If the same constant force brings both cars to a stop, how far will

car B travel in coming to rest as compared to the distance car A traveled
Physics
1 answer:
evablogger [386]3 years ago
7 0

Answer:four times

Explanation:

Given

mass of both cars A and B are same suppose m

but velocity of car B is same as of car A

Suppose velocity of car A is u

Velocity of car B is 2 u

A constant force is applied on both the cars such that they come to rest by travelling certain distance

using  to find the distance traveled

where, v=final velocity

u=initial velocity

a=acceleration(offered by force)

s=displacement

final velocity is zero

For car A

0-(u)^2=2\times a\times s

s_a=\frac{u^2}{2a}------1

For car B

0-(2u)^2=2\times a\times s_b

s_b=\frac{4u^2}{2a}----2

divide 1 and 2 we get

\frac{s_a}{s_b}=\frac{1}{4}

thus s_b=4\cdot s_a

distance traveled by car B is four time of car A

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A 50 kg astronaut floating in space throws her 2 kg wrench to the left at 10
Hoochie [10]

The astronaut will move at 0.4 m/s in the opposite direction to the wrench

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the astronaut-wrench system must be conserved before and after the launch.

Before the launch, the total momentum is zero, since the astronaut is at rest:

p = 0 (1)

After the launch, the total momentum is:

p=mv+MV (2)

where :

m = 2 kg is the mass of the wrench

v = 10 m/s is the velocity of the wrench

M = 50 kg is the mass of the astronaut

V is the recoil velocity of the astronaut

Since momentum is conserved, we can write (1) = (2), and so we can solve for V:

0=mv+MV\\V=-\frac{mv}{M}=-\frac{(2)(10)}{50}=-0.4 m/s

And the negative sign means that the astronaut will move in the opposite direction to the wrench.

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3 years ago
In an engine governor, the two spheres (total mass of 1.0kg) are at 0.05m and rotating at 37rad/s If the engine increases the an
kirill115 [55]

Here we can say that there is no external torque on this system

So here we can say that angular momentum is conserved

so here we will have

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now we have

I_1 = mr^2

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similarly let the final distance is "r"

so now we have

I_2 = mr^2

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4 0
3 years ago
Read 2 more answers
If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?
Law Incorporation [45]

a) 4.8 m/s² his average acceleration during the first 2.5 s

b) 15 m distance he cover during the first 2.5 seconds.

c) 31 m/s his speed as he finished the race.

<h3>(a) How to calculate the Average Acceleration ?</h3>

Average acceleration is calculated by using Newtons first law of motion

v = u + at

where

u is initial velocity

v is final velocity

a is constant acceleration

t is time

Given u = 0m/s , v = 12 m/s and t = 2.5 sec

Therefore average acceleration is given by

a=\frac{v}{t}

a=\frac{12}{2.5}

a = 4.8 m/sec²

b) Here we will use newtons second law of motion

s=ut+\frac{1}{2}at^{2}

On substituting value we get

s = 15m

c) Here we will use newtons third law of motion

v² = u² + 2as

Here u = 0 , a = 4.8 m/s² and s = 100 m

Therefore

v² = 960

v = 31 m/s

Disclaimer: the question was given incomplete in the portal. Here is the complete question.

Question: Usain Bolt's 100m sprint Runner set the world record for the100 meter sprint during the 2009World Championships in Berlin with a time of 9.58 s, reaching a top speed of 12 m/s in about 2.5 s.

a. What was his average acceleration during the first 2.5 s?

b. What distance did he cover during the first 2.5 seconds, assuming his acceleration was constant?

c. If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

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