Answer:
Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .
Explanation:
The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.
1)

<Solve using the formula which is:
Mass=Density×Volume

2)
3)

(Length × Width × Height)


(Answer=7.5)
Answer: 1477.78 N
Explanation:
Let's assume that the cross sectional area of the smaller piston be A1
let's also assume the cross sectional area of the larger piston be A2
We assume the force applied to the smaller piston be F1
We also assume the force applied to the larger piston be F2
we then use the formula
F1/A1 = F2/A2
From our question,
The radius of the smaller piston is 5 cm = 0.05 m
The radius of the larger piston is 15 cm = 0.15 m
The force of the larger piston is 13300 N
The force of the smaller piston is unknown = F
A1 = πr² = 3.142 * 0.05² = 0.007855 m²
A2 = πr² = 3.142 * 0.15² = 0.070695 m²
F1/0.007855 = 13300/0.070695
F1 = (13300 * 0.007855) / 0.070695
F1 = 104.4715 / 0.070695
F1 = 1477.78 N
Thus, the force the compressed air must exert is 1477.78 N
Answer:
The frequency of these waves is 
Explanation:
Given that,
Wavelength = 6.6 km
Distance = 8810 km
Time t = 8.67 hr
We need to calculate the velocity of sound
Using formula of velocity

Where, D = distance
T = time
Put the value into the formula


We need to calculate the frequency
Using formula of frequency


Put the value into the formula





Hence, The frequency of these waves is 