Answer:
h = v₀² / 2g
, h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²
<u>Answer:</u>
Cannonball will be in flight before it hits the ground for 2.02 seconds
<u>Explanation:</u>
Initial height from ground = 20 meter.
We have equation of motion , 
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 
, we need to calculate time when s = 20 meter.
Substituting
So it will take 2.02 seconds to reach ground.
Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ
R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m
