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3 years ago

7

A ball rolling along a floor doesn't continue rolling indefinitely. is it because it is seeking a place of rest or because some

force is acting upon it?

1 answer:

Ivan3 years ago

7 0

It is because gravity is pushing down on the ball

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PLEASEEEE HEEEEELP!!!!!

8090 [49]

Answer:

Before:

$p_{truck}=16400\ kg.m/s$

$p_{car}=10000\ kg.m/s$

After:

$p_{truck}=8000\ kg.m/s$

$p_{car}=8400\ kg.m/s$

$v_{fcar}=8.4\ m/s$

$F=9333.33 \ Nw$

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

$p_1=m_1v_{1o}+m_2v_{2o}$

After the collision, they have speeds of v1f and v2f and the total momentum is

$p_2=m_1v_{1f}+m_2v_{2f}$

Impulse J is defined as

$J=F.t$

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

$J=\Delta p$

As the total momentum is conserved:

$p_1=p_2$

$m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}$

We can compute the speed of the second object by solving the above equation for v2f

$\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{ m_2 }$

The given data is

$m_1=2000\ kg$

$m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s$

a) The impulse will be computed at the very end of the answer

b) Before the collision

$p_{truck}=2000\cdot 8.2=16400\ kg.m/s$

$p_{car}=1000\cdot 0=0\ kg.m/s$

c) After collision

$p_{truck}=2000\cdot 4=8000\ kg.m/s$

Compute the car's speed:

$\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }$

$v_{2f}=8.4\ m/s$

And the car's momentum is

$p_{car}=1000\cdot 8.4=8400\ kg.m/s$

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

$J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s$

The force can be computed as

$\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw$

The force on the car has the same magnitude and opposite sign

7 0

4 years ago

What is the mass of a crate if a force of 200 N causes it to accelerate at 8 m/s2? (Formula: F=ma)

Ugo [173]

25 kg

Explanation:

we can calculate the mass of the crate by using the Newton's second law:

$F=ma$

where

F is the force acting on the crate

m is the mass

a is the acceleration

In this problem, $F=200 N$ and $a=8 m/s^2$ , so we can re-arrange the equation above to calculate the mass of the crate:

$m=\frac{F}{a}=\frac{200 N}{8 m/s^2}=25 kg$

4 1

3 years ago

Read 2 more answers

a dad expends 200 joules of work pushing a stroller up a hill. he produces 125 j of work. how efficient is he being?

AVprozaik [17]

Not efficient enough

7 0

3 years ago

A rectangular certificate has a perimeter of 32 inches. Its area is 63 square inches. What are the dimensions of the certificate

irina1246 [14]

Perimeter = 2 ( L + W )
32 = 2 ( L + W )
16 = L + W
L = 16 - W

Area = L W
63 = L W
63 = (16-W) W
63 = 16W - W²
-W² + 16 W - 63 = 0
By factorizing W = 9 or W = 7
So the dimensions are 7 and 9

3 0

3 years ago

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

$15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm$

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

5 0

3 years ago