Answer:
2.210N
Explanation:
Workdone = Force x distance
Distance = 38m , Workdone = 84J
Hence 84J = Force x 38m
Force = 84J / 38m
Force = 2.210N =2.2N
Given that:
k = 500 n/m,
work (W) = 704 J
spring extension (x) = ?
we know that,
Work = (1/2) k x²
704 = (1/2) × 500 × x²
x = 1.67 m
A spring stretched for 1.67 m distance.
Answer:
spring deflection is x = (v2 / R + g) m / 4
Explanation:
We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration
Let's write the equations on the Y axis of this description
Fe - W = m 
Where Fe is elastic force, W the weight and
the centripetal acceleration. The elastic force equation is
Fe = - k x
4 (k x) - mg = m v² / R
The four is because there are four springs, R is theradio of dip
We can calculate the deflection (x) of the springs
x = (m v2 / R + mg) / 4
x = (v2 / R + g) m / 4
Answer:
As we need to use a nested loop in our function,hence push $ra
pop $ra
jal nested_function_label
nop is the correct option.
Answer:
The value is 
Explanation:
From the question we are told that
The mass of the car is
The period of the circular motion is 
The radius is 
Generally the frequency of the circular motion is

=> 
=> 