<em><u>1.car</u></em><em><u> </u></em><em><u>towing</u></em>
<em><u>2.pulling</u></em><em><u> </u></em><em><u>bucket</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>water</u></em>
<em><u>3.gym</u></em><em><u> </u></em><em><u>equipment</u></em><em><u> </u></em>
<em><u>4.crane</u></em><em><u> </u></em><em><u>machine</u></em>
<em><u>5.tug</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>war</u></em>
This can be seen as a trick question because heat engines can typically never be 100 percent efficient. This is due to the presence of inefficiencies such as friction and heat loss to the environment. Even the best heat engines can only go up to around 50% efficiency.
Answer:
θ=108rad
t =10.29seconds
α=-8.17rad/s²
Explanation:
Given that
At t=0, Wo=24rad/sec
Constant angular acceleration =30rad/s²
At t=2, θ=432rad as it try to stop because the circuit break
Angular motion
W=Wo+αt
θ=Wot+1/2αt²
W²=Wo²+2αθ
We need to find θ between 0sec to 2sec when the wheel stop
a. θ=Wot+1/2αt²
θ=24×2+1/2×30×2²
θ=48+60
θ=108rad.
b. W=Wo+αt
W=24+30×2
W=84rad/s
This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.
Wo=84rad/sec
W=0rad/s, because the wheel stop at θ=432rad
Using W²=Wo²+2αθ
0²=84²+2×α×432
-84²=864α
α=-8.17rad/s²
It is negative because it is decelerating
Now, time taken for the wheel to stop
W=Wo+αt
0=84-8.17t
-84=-8.17t
Then t =10.29seconds.
a. θ=108rad
b. t =10.29seconds
c. α=-8.17rad/s²
Answer: Temperature in Fahrenheit is 577.4
Explanation:
The conversion factor for converting celcius to Fahrenheit is:

where F = temperature in Fahrenheit
C = Temperature in Celcius
Given : Temperature difference in Celcius = 
Putting in the values we get:


Thus the answer in Fahrenheit is 577.4
Answer:
The speed should be reduced by 1/√2 or 0.707 times
Explanation:
The relationship between the kinetic energy, mass and velocity can be represented by the following equation:
K.E = ½m.v²
In this equation, the mass is inversely proportional to the square of the velocity or speed. This means that as the mass increases, the speed reduces by × 2.
Let; initial mass = m1
Final mass = m2
Initial velocity = v1
Final velocity = v2
According to the question, if the mass of the body is doubled i.e. m2 = 2m
½m1v1² = ½m2v2²
½ × m × v1² = ½ × 2m × v2²
Multiply both sides by 2
(½ × m × v1²)2 = (½ × 2m × v2²)2
m × v1² = 2m × v2²
Divide both sides by m
v1² = 2v2²
Divide both sides by 2
v1²/2= v2²
Square root both sides
√v1²/2= √v2²
v1/√2 = v2
v2 = 1/√2 v1
This shows that to maintain the same kinetic energy if the mass is doubled, the speed should be reduced by 1/√2 or 0.707 times.