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ehidna [41]
2 years ago
6

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

ilograms. What is the kinetic energy of the system after the collision if the collision is elastic?
Physics
1 answer:
rewona [7]2 years ago
3 0

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

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Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop
anastassius [24]
To calculate the acceleration of the wooden block, we use the expression F=ma where F is the force applied, m is the mass of the object and a is the acceleration. We calculate as follows:

F = ma
4.9 = 0.5a
a = 9.8 

Hope this answers the question. Have a nice day.

4 0
2 years ago
Read 2 more answers
A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular speed of the merry-go-round after the person jumps on?

We can apply The Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0
2 years ago
If you drive through water, your brakes may become slippery and ineffective. To dry the brakes off, __________.
GenaCL600 [577]

Answer:

I am not a driver, but I think it's C.

Explanation:

6 0
2 years ago
Read 2 more answers
two people are standing 100m apart on the bank of a river that flows due east. if a rock on the opposite bank is along a bearing
salantis [7]

Answer:160.88 m

Explanation:

Given

Distance Between two person is 100

i.e. p+q=100

Let W be the width of River

From First Triangle

p=W\tan 24

From Second Triangle

q=W\tan 10

W\tan 24+W\tan 10=100

W(\tan 24+\tan 10)=100

W=\frac{100}{0.6215}

W=160.88 m

       

3 0
2 years ago
A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C
Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 \frac{J}{kg k}

The heat of vaporization is =  2256 \frac{KJ}{kg}

Mass = 0.030 kg

Energy needed to heat water to boiling point =  m c ( T_{2} - T_{1} )

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy (E_{1}) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

E_{2} = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E =  E_{1} + E_{2}

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

9 0
2 years ago
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