By definition we have to:
Applied force: It is the external force that acts directly on a body.
Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.
This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.
Answer:
you exert an Applied Force on an object when you pull it towards you
A. Applied Force
Answer:
(a) 
(b) 
(c) K.E. = 21.168 J
(d) 
Explanation:
Given:
- mass of a block, M = 3.6 kg
- initial velocity of the block,

- constant downward acceleration,

That a constant upward acceleration of
is applied in the presence of gravity.
∴
- height through which the block falls, d = 4.2 m
(a)
Force by the cord on the block,



∴Work by the cord on the block,


We take -ve sign because the direction of force and the displacement are opposite to each other.

(b)
Force on the block due to gravity:

∵the gravity is naturally a constant and we cannot change it


∴Work by the gravity on the block,



(c)
Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.
mathematically:


K.E. = 21.168 J
(d)
From the equation of motion:

putting the respective values:

is the speed when the block has fallen 4.2 meters.
Answer:
-1.5m/s²
Explanation:
Acceleration can be thought of as [Change in Velocity]/[Change in time]. To find these changes, you simply subtract the initial quantity from the final quantity.
So for this question you have:
- V_i = 110m/s
- V_f = 80m/s
- t_i = 0s
- t_f = 20s
which means that the acceleration = (80-110)/(20-0)[m/s²] = (-30/20)m/s² = -1.5m/s²
Answer:
The first frequency of audible sound in the speed sound is
f = 662 Hz
Explanation:
vs = 344 m/s
x = 52 cm * 1 / 100m = 0.52m
The wave length is the distance between the peak and peak so
d = 2x
d = 2*0.52 m
d = 1.04 m
So the frequency in the speed velocity is
f = 1 / T
f = vs / x = 344 m/s / 0.52m
f ≅ 662 Hz
I think it is D. I hope this helps