To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

Here,
k = Spring constant
m = Mass
Our values are given as,


Rearranging to find the spring constant we have that,




Therefore the spring constant is 1.38N/m
Explanation:
Mass of the ball, m = 0.058 kg
Initial speed of the ball, u = 11 m/s
Final speed of the ball, v = -11 m/s (negative as it rebounds)
Time, t = 2.1 s
(a) Let F is the average force exerted on the wall. It is given by :


F = 0.607 N
(b) Area of wall, 
Let P is the average pressure on that area. It is given by :


P = 0.202 Pa
Hence, this is the required solution.
I'm assuming the question is what is the robin's speed relative to to the ground...
Create an equation that describes its relative motion.
rVg = rVa + aVg
Substitute values.
rVg = 12 m/s [N] + 6.8 m/s [E]
Use vector addition.
| rVg | = √ | rVa |² + | aVg |²
| rVg | = √ 144 m²/s² + 46.24 m²/s²
| rVg | = √ 19<u>0</u>.24 m²/s²
| rVg | = 1<u>3</u>.78 m/s
Find direction.
tanФ = aVg / rVa
tanФ = 6.8 m/s / 12 m/s
Ф = 29°
Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]
Answer:
= 33.33 cm
Explanation:
Given:
When mass,
=21 kg
distance travelled is
= 140 cm
When mass,
=5 kg
distance travelled is
= ?
Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.
Therefore according to the question,


= 33.33 cm
Distance travelled is 33.33 cm when mass is 5 kg.
Explanation:
Let us assume that the separation of plate be equal to d and the area of plates is
. As the capacitance of capacitor is given as follows.
C = 
It is known that the dielectric strength of air is as follows.
E = 
Expression for maximum potential difference is that the capacitor can with stand is as follows.
dV = E × d
And, maximum charge that can be placed on the capacitor is as follows.
Q = CV
= 
= 
= 
= 
or, = 10.62 nC
Thus, we can conclude that charge on capacitor is 10.62 nC.