Answer:
<em>The balloon is 66.62 m high</em>
Explanation:
<u>Combined Motion
</u>
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is
The values are
We must find the values of t such that the height of the camera is 0 (when it hits the ground)
Multiplying by 2
Clearing the coefficient of 
Plugging in the given values, we reach to a second-degree equation
The equation has two roots, but we only keep the positive root
Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is
There's an effect called The Doppler Effect, that's the "up and down" sound.
That frequency is 1.8mm and 115 per minute.
(Hope this helped.)
Answer:
Work done, W = 19.6 J
Explanation:
It is given that,
Mass of the block, m = 5 kg
Speed of the block, v = 10 m/s
The coefficient of kinetic friction between the block and the rough section is 0.2
Distance covered by the block, d = 2 m
As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.
W = 19.6 J
So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.
Newton is a unit of force. Joules is an amount of work which is equal to the force times distance, or newton meters. So the product of newtons and meters makes joules.
Answer:
4 seconds
Explanation:
Average acceleration is change in velocity over change in time:
a = Δv / Δt
Δt = Δv / a
Δt = (15 m/s - 5 m/s) / 2.5 m/s²
Δt = 4 s
