If my calculations are accurate, the angle is 67.5 degrees.
The force exerted on the car during the stop measures 6975 N.
Explanation: Given that the mass (m) is 930 kg, speed (s) at 56 km/h converts to 15 m/s, and the stopping time (t) is 2 s, we compute the force using F = m * a. Here, acceleration (a) can be obtained through a = s/t. The total force calculation confirms that F = 930 kg * (15 m/s) / 2 s results in 6975 N.
Response:
The ball remained airborne for 3.896 seconds
Explanation:
Given that
g = 9.8 m/s², representing gravitational acceleration,
If the angle of launch is 45°, the horizontal range will be maximized.
Both horizontal and vertical launch velocities are equal, each equating to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The duration to reach maximum height is half of the flight time.
v = u + at ∵ v = 0 (at maximum height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The total time in the air equals twice the time to reach maximum height
2 t₁ = 3.896 s
The horizontal distance covered is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in the air for 3.896 seconds
I don't know that; sorry, I should just be removed from here.
Answer:
The kinetic energy is higher for the first cart.
Explanation:
For the second cart, its mass is 2kg and the momentum measured is 10kg m/s, which leads to
resulting in
.
Consequently, the kinetic energy for the 3kg cart ends up as
indicating it is less than that of the 1kg cart so it follows that the first cart possesses greater kinetic energy.
