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Genrish500 [490]
3 years ago
5

A plastic ball of mass 0.200 kg moves with a velocity of 0.30 m/s. This plastic ball collides with a second plastic ball of mass

0.100 kg that is moving along the same line at a velocity of 0.10 m/s. After the collision, the velocity of the 0.100 kg ball is 0.36 m/s. What is the velocity of the second ball?
Physics
1 answer:
GrogVix [38]3 years ago
8 0
<h2>Velocity of 0.2 kg ball is 0.17 m/s</h2>

Explanation:

Let the mass of balls be m and M.

Initial velocity ball 1 be u₁, ball 2 be u₂

Final velocity ball 1 be v₁, ball 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 0.2 x 0.3 + 0.1 x 0.1 = 0.07 kgm/s

Final momentum = m x v₁ + M x v₂ = 0.2 x v₁ + 0.1 x 0.36 = 0.2v₁ + 0.036

We have momentum conservation

Initial momentum = Final momentum

0.07 = 0.2v₁ + 0.036

v₁ = 0.17 m/s

Velocity of 0.2 kg ball is 0.17 m/s

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You have been called in by your neighborhood safety council to advise them about installing a barrier on a dangerous curve that
Alborosie

Answer: The softer barrier is the better option

Explanation:

1) When is a car is moving at a certain speed, it has a certain amount of momentum (p=mv). A collision against a barrier would cause its momentum to decrease to 0. A change in momentum is Impulse

2) The formula for Impulse: J = f * Δt

J is Impulse

f is the force applied during the time Δt

A tough barrier would produce a smaller Δt, which means more force is applied on the car. (J is always constant)

A softer barrier would apply less force on the car, which means Δt is large.

Answer: The softer barrier is the better option

4 0
3 years ago
A person drives north 3 blocks, then turns east and drives 3 blocks. The driver then turns south and drives 3 blocks. How could
KonstantinChe [14]

Answer:by driving east 3 blocks from the starting point

Explanation:)

4 0
2 years ago
A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How
Nata [24]
Hello!

Answer:

2000 J

Explanation

Work equation is expressed as:

W=F.d.Cos \alpha

Where:

F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)

By substituting:

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Have a nice day!
8 0
3 years ago
Can someone please help?
Diano4ka-milaya [45]

I think you go to my school!

7 0
3 years ago
Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L
snow_lady [41]

Answer:

Q=81.56\ W/m^2

Explanation:

Given that

T_1= 210 K

T_2= 150 K

Emissivity of surfaces(∈) = 1

We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

Q=\sigma (T_1^4-T_2^4)\ W/m^2

So now by putting the values

Q=\sigma (T_1^4-T_2^4)\ W/m^2

Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2

Q=81.56\ W/m^2

So rate of heat transfer per unit area

Q=81.56\ W/m^2

8 0
3 years ago
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