Answer:
44 N/m
Explanation:
The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m
The work needed to stretch a spring by <em>e</em> is given by
where <em>k</em> is spring constant.
Using the appropriate values,
Incomplete question.The complete question is here
What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.
Answer:
F= 0.034 N
Explanation:
Given Data
Outer=2 cm
Edge of blade=21 cm
Mass=7.7 g
Length of blade=21 cm
The last 2cm is treated as if they were at a point 20cm from the center of rotation
To Find
Force=?
Solution
Convert the given frequency to angular frequency
ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)
ω= 3/2*π rad/sec
Now to find centripetal force.
F = m×v²/r
F= m×ω²×r
Put the data
F = 0.0077 kg × (3/2×π rad/sec
)²× 0.20 m
F= 0.034 N
The answer is :
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Answer:
Option d)
Solution:
As per the question:
Work done by farm hand, 
Force exerted, F' = 310 N
Angle, 
Now,
The component of force acting horizontally is F'cos
Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.
Thus
d = 3.406 m = 3.4 m
