Answer : B. 0 m/s∧2
Now we know that velocity = distance/ time
And acceleration = velocity/time.
From the graph we see that at the time interval 2 secs,the distance travelled is 10 m. Hence the velocity(v1) =10/2 = 5m/s.
From the graph we see that at the time interval 5 secs,the distance travelled is 25 m. Hence the velocity (v2) =25/5 = 5m/s.
Now acceleceration between the time interval of 2 ad 5 secs is calculated as
(v1-v2)/t2-t1 = (5-5)/(5-2) = 0 m/s∧2
D............................
Answer:
49.5 Hz.
Explanation:
From the question given above, the following data were obtained:
Period (T) = 0.0202 s
Frequency (f) =?
The frequency and period of a wave are related according to the following equation:
Frequency (f) = 1 / period (T)
f = 1/T
With the above formula, we can obtain the frequency of the wave as follow:
Period (T) = 0.0202 s
Frequency (f) =?
f = 1/T
f = 1/0.0202
f = 49.5 Hz
Therefore the frequency of the wave is 49.5 Hz.
a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:

Therefore the work done is 17673 J.
b)
The power is given by:

the problem states that the time it takes is 6 s. Then:

Therefore the power is 2945.5 W