Answer:
F₂= 210 pounds
Explanation:
Conceptual analysis
Hooke's law
Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:
F= K*x Formula (1)
Where;
F is the magnitude of the force applied to the spring in Newtons (Pounds)
K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)
x the elongation of the spring (inch)
Data
The data given is incorrect because if we apply them the answer would be illogical.
The correct data are as follows:
F₁ =80 pounds
x₁= 8 inches
x₂= 21 inches
Problem development
We replace data in formula 1 to calculate K :
F₁= K*x₁
K=( F₁) / (x₁)
K=( 80) / (8) = 10 pounds/ inche
We apply The formula 1 to calculate F₂
F₂= K*x₂
F₂= (10)*(21)
F₂= 210 pounds
The work done on the box by the applied force is zero.
The work done by the force of gravity is 75.95 J
The work done on the box by the normal force is 75.95 J.
<h3>The given parameters:</h3>
- Mass of the box, m = 3.1 kg
- Distance moved by the box, d = 2.5 m
- Coefficient of friction, = 0.35
- Inclination of the force, θ = 30⁰
<h3>What is work - done?</h3>
- Work is said to be done when the applied force moves an object to a certain distance
The work done on the box by the applied force is calculated as;
where;
a is the acceleration of the box
The acceleration of the box is zero since the box moved at a constant speed.
The work done by the force of gravity is calculated as follows;
The work done on the box by the normal force is calculated as follows;
Learn more about work done here: brainly.com/question/8119756
Answer:
r = 0.02 m
Explanation:
from the question we have :
speed = 1 rps = 1x 60 = 60 rpm
coefficient of friction (μ) = 0.1
acceleration due to gravity (g) = 9.8 m/s^{2}
maximum distance without falling off (r) = ?
to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force
mv^2 / r = m x g x μ
v^2 / r = g x μ .......equation 1
where
velocity (v) = angular speed (rads/seconds) x radius
angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm
angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds
now
velocity = 6.28 x r = 6.28 r
now substituting the value of velocity into equation 1
v^2 / r = g x μ
(6.28r)^2 / r = 9.8 x 0.1
39.5 x r = 0.98
r = 0.02 m
Answer:
Total work done in expansion will be 
Explanation:
We have given pressure P = 2.10 atm
We know that 1 atm 
So 2.10 atm 
Volume is increases from 3370 liter to 5.40 liter
So initial volume 
And final volume 
So change in volume 
For isobaric process work done is equal to 
So total work done in expansion will be
