Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.

The maximum velocity is

Answer:
The curl is 
Explanation:
Given the vector function

We can calculate the curl using the definition

Thus for the exercise we will have

So we will get

Working with the partial derivatives we get the curl

As the length increases, resistance increases, as a result current decreases.
The average speed is the distance per time ratio. velocity is the rate at which the position changes
Answer:
A. 1.172 metres
B. 6.82 Ns
C. 4.796 m/s
Explanation:
The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.
The 1.40 kg block is at rest so velocity is zero and has no momentum.
The bullet of mass 22 g = 0.022 kg with velocity of 310 m/s
Momentum = 310*0.022
Momentum = 6.82 Ns.
If the bullet gets embedded they will both have common velocity v
6.82 = (0.022+1.40)v
6.82 = 1.422v
V = 6.82/1.422
V = 4.796 m/s
How high the block will rise after the bullet is embedded is given by
H = (U²Sin²tita)/2g
Where tita is 90°
H = (4.796² * sin²(90))/(2*9.81)
H =( 23.001616*1)/19.62
H = 1.172 metres