The work done when a spring is stretched from 0 to 40cm is 4J.
What is work done?
Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.
The work done on the spring to stretch to 40cm is,
F = kx
where F is force, k is force constant.
k = F / x = 10 N / 20 * 10^-2 m = 50 N/m
W = 0.5 * k * (x)^2
where W = work done, k = force constant.
W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.
Therefore, the work done on the spring when it is stretched to 40cm is 4J.
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Answer:
0 m/s
Explanation:
Relative speed is defined as the speed of an object with respect with another object.
In other words, it is the speed of an object as viewed from the frame of reference of another object.
When two objects are moving in the same direction, their relative velocity is given as:
where velocity of first object
velocity of second object
In the case of the two cars, 22 m/s
Therefore:
Their relative velocity is 0 m/s.
Answer: 215.15 N
Explanation:
If we draw a free body diagram of the mass we will have the following:
(1)
(2)
Where is the tension force of the rope, the mass, the acceleration due gravity and is the weight.
On the other hand, we can calculate as follows:
Where and
(3)
Now, we firstly need to find from (2):
(4)
(5)
Substituting (5) in (1):
(6)
Finally:
Answer:
to stretch their muscles so they don't strain or hurt themselves
Explanation:
hi!
Answer:
The reason is because the pressure of the air inside the room drops with time which makes opening the door to require an increased amount of force to make up for the reduced pressure inside the room
Explanation:
From the kinetic theory of gases we have the following relation;
Where:
K = Boltzmann constant
T = Temperature
m = Mass
MW = Molecular weight
V = Volume
= Root mean square velocity
Whereby the room door is closed, the kinetic energy of the air particles will be used up such that the average velocity of the particles will decrease, given that the volume of the room is constant, the pressure inside the room will drop below the original pressure outside the room such that the force on the door due to the outside pressure is larger than the force on the door from inside the room requiring a larger amount of force to overcome the resistance of the now higher outside pressure force.