Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
Answer:
Explanation:
This problem can be solved with the conservation of the momentum.
If the ball is fired upward, the momentum before and after the ball is fired must conserve. Hence, the speed of the ball is the same that the speed of the car just in the moment in wich the ball is fired.
Hence, the result depends of the acceleration of the car. If the change in the speed is higher than the speed of the ball, it is probably that the ball will be behind the car or it will come back to the car.
If the ball is fired forward, and if the change in the speed of the car is not enogh, the ball will be in front of the car.
HOPE THIS HELPS!!
Answer:
1) True, 2) True, 3) False, 4) False, 5) False
Explanation:
1) True. Dissipative energy cannot be recovered, in general it is a form of heat
2) True. The dissipation can be by radiation, heat
3) False. Mechanical energy is divided into K and U but not in equal parts
4) False. When there are dissipative interactions, part of the mechanical energy is set in the form of heat, so its value decreases
5) False. Mechanical energy is the sum of those two energies
A. scatter plot (100% positive, btw quizzes is a good tool for answers)
Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately
