Answer:
10 m/s
Explanation:
Momentum before collision = momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v
64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v
60 kg m/s = (6 kg) v
v = 10 m/s
Answer:
You increase the acceleration of the car.
Answer:
4.45×10¯¹¹ N
Explanation:
From the question given above, the following data were obtained:
Mass of ball (M₁) = 4 Kg
Mass of bowling pin (M₂) = 1.5 Kg
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Distance apart (r) = 3 m
Force of attraction (F) =?
The force of attraction between the ball and the bowling pin can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 4 × 1.5 / 3²
F = 4.002×10¯¹⁰ / 9
F = 4.45×10¯¹¹ N
Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N
Answer:
C.) 1.5 kg
Explanation:
Start with the equation:

Plug in what you know, and solve:

Find matching soluation:
C.) 1.5 kg
Answer:

Explanation:
<u>Elastic Potential Energy
</u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.
Solving for Δx:

Substituting:

Calculating:

