Answer:
50
Explanation:
The mechanical advantage of a machine is given by

where
is the output force
is the input force
For the crowbar in this problem,
is the force in input applied by the worker
is the force that the machine must apply in output to overcome the resistance of the window and to open it
Substituting into the equation, we find

Answer: A) mass on earth surface = 5.91kg
B) mass on surface of jupiter = 5.91kg
C) weight on surface of jupiter = 10.697N
Explanation:
The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below
W=mg
From the question, g= 9.8m/s² and weight on the surface on the earth is 58N
A) The mass of watermelon on earth is
m = 58/ 9.8 = 5.91kg
B) the mass of the watermelon on jupiter is 5.91kg.
You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.
C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as
W = mg
W = 5.91 x 9.8
= 10.697N.
The frequency of oscillation is 2.153 Hz
What is the frequency of spring?
Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.
For the mass-spring system in this problem,
The Frequency of spring is calculated with the equation:

Where,
f = frequency of spring
k = spring constant = 64 N/m
m = mass attached to spring = 350g = 0.350 kg
a = maximum acceleration = 5.3 m/s^2
Substituting the values in the equation,



Hence,
The frequency of oscillation is 2.153 Hz
Learn more about frequency here:
<u>brainly.com/question/13978015</u>
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Answer:
The equation of motion is 

Explanation:
Lets calculate
The weight attached to the spring is 24 pounds
Acceleration due to gravity is 
Assume x , is spring stretched length is ,4 inches
Converting the length inches into feet 
The weight (W=mg) is balanced by restoring force ks at equilibrium position
mg=kx
⇒ 
The spring constant , 
= 72
If the mass is displaced from its equilibrium position by an amount x, then the differential equation is



Auxiliary equation is, 

=
Thus , the solution is 

The mass is released from the rest x'(0) = 0
=0


Therefore ,

Since , the mass is released from the rest from 4 inches
inches
feet
feet
Therefore , the equation of motion is 