Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.
We need to find its height.
We will use the formula P.E = mgh
Therefore h = P.E / mg
where P.E is the potential energy,
m is mass in kg,
g is acceleration due to gravity (9.8 m/s²)
h is the height of the object's displacement in meters.
h = P.E. / mg
h = 14000 / 40 × 9.8
h = 14000 / 392
h = 35.7
Therefore the canon ball was 35.7 meters high.
The distance from a 27 mw point source of electromagnetic waves where the electric field amplitude 0. 060 v/m will be 21.21 m .
Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields.
The highest point of a wave is known as 'crest' , whereas the lowest point is known as 'trough'. Electromagnetic waves can be split into a range of frequencies. This is known as the electromagnetic spectrum.
c = 3 * 
m/s
∈ = 8.85 * 
/ N/ 
E = 0. 060 v/m
I = P / 4π
Also , I = c ∈ 
/2
= P / 4π I equation 1
substituting the value of I in equation 1
= 2 P / 4π (c ∈ )
= 2 * 27 * 
/ 4 * 3.14 * 3 * * 8.85 * * 
r = 21.21 m
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C according to my calculations
The car takes 10.8 seconds to reach a velocity of 42.1 m/s
Explanation:
The motion of the car is at constant acceleration, so we can use the following suvat equation:
where
v is the final velocity of the car
u is its initial velocity
a is the acceleration
t is the time
For the car in this problem, we have:
u = 15.0 m/s
v = 42.1 m/s
Solving for t, we find the time it takes for the car to reach that velocity:
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