To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.
Where
A = Cross-sectional Area
Y = Young's modulus
= Coefficient of linear expansion for steel
= Temperature Raise
Our values are given as,
Replacing we have,
Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N
<span>First, we use the kinetic energy equation to create a formula:
Ka = 2Kb
1/2(ma*Va^2) = 2(1/2(mb*Vb^2))
The 1/2 of the right gets cancelled by the 2 left of the bracket so:
1/2(ma*Va^2) = mb*Vb^2 (1)
By the definiton of momentum we can say:
ma*Va = mb*Vb
And with some algebra:
Vb = (ma*Va)/mb (2)
Substituting (2) into (1), we have:
1/2(ma*Va^2) = mb*((ma*Va)/mb)^2
Then:
1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2
We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1:
1/2(ma) = (ma^2)/mb
Cancel the ma of the left, leaving the right one with exponent 1:
1/2 = ma/mb
And finally we have that:
mb/2 = ma
mb = 2ma</span>
Answer:
a=
Explanation:
The net force, 
of the box is expressed as a product of acceleration and mass hence
where m is mass and a is acceleration
Making a the subject, a= 
From the attached sketch,
∑ 
where 
is frictional force and 
is horizontal angle
Substituting ∑ as in the equation where we made a the subject
a= 
Since we’re given the value of F as 240N, as 41.5N, as 
and mass m as 30kg
a= 
Remain the same
Explanation:
If the force exerted by the intern is doubled and the distance is halved, the work done by the intern remains the same.
Work done is the force applied to move a body through a distance.
Work done = F x d
where F is the applied force
d is the distance moved
Now;
if:
f = 2f
d = 
d
Input the parameter:
Work done = fxd = 2f x d = fd
The work done will still remain the same
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Answer:
F₁ = F₂ = F₃ = 0 N
Explanation:
given,
Arrow 1 mass = 80 g speed = 10 m/s
Arrow 2 mass = 80 g speed = 9 m/s
Arrow 3 mass = 90 g speed = 9 m/s
Horizontal Force:- F₁ , F₂ and F₃
There is no air resistance.
If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.
We know,
According to newton's second law
F = m a
If Acceleration is equal to zero
Then Force is also equal to zero.
Hence, F₁ = F₂ = F₃ = 0 N
