If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
Answer:
speaker64
--------
34x
Explanation:
64-34
x
speaker
4
2
4
788
- circuit
voltage
100000
x.34
Sorry but you have no picture shown
Answer:
Explanation:
Given that,
The compression in the spring, x = 0.0647 m
Speed of the object, v = 2.08 m/s
To find,
Angular frequency of the object.
Solution,
We know that the elation between the amplitude and the angular frequency in SHM is given by :
A is the amplitude
In case of spring the compression in the spring is equal to its amplitude
So, the angular frequency of the spring is 32.14 rad/s.
Answer:
1.#potential energy = PE, m = mass in kg, g = force of gravity, h= vertical height above the ground. ** means to the power of ie exponent. * means multiply.
PE = mgh
300 = m(10)(15)
m = 300/(10)(15)
m= 2kg
2. KE = 1/2 mv**2
= 1/2(50)(50)**2
= 2500 joules
Explanation
Is as in solution
