Answer:
a)Work done by fireman= 2.15 Btu
b) Time t= 0.86 sec
Explanation:
Given that
Weight = 280 lbf
We know that 1 lbf = 4.44 N
so 280 lbf = 1245.5 N
Weight =1245.5 N
Height h = 60 ft
We know that
1 ft = 0.3048 m
So 60 ft = 18.28 m
h =18.28 m
Power = 3.5 hp
We know that
1 hp =0.74 KW
So 3.5 hp = 2.61 KW
Power = 2.61 KJ/s
So the work done by fireman = Weight x h
Now by putting the values
Work done by fireman= 1245.5 x 18.28 J
Work done by fireman= 2267.74 J
Work done by fireman= 2.26774 KJ
We know that 1 Btu= 1.05 KJ
So 2.266 KJ = 2.15 Btu
Work done by fireman= 2.15 Btu
We know that ,rate of work is called power.
Power x time = work
2.61 x t = 2.26
So t= 0.86 sec
It is symbols to make it easier and to save space!
<span>When Kevin pulls his cotton shirt off his body, the electrons get transferred from the shirt (in form of static charges i.e. electrons to the body. So, the shirt becomes positively charged and Kevin’s body becomes negatively charged.
As a result of charge transfer from the shirt to the body, we can hear a crackling sound. or if observed in dark, a sparkle can be seen.</span>
We want to know what is the power supplied by the power cell if the current I=0.5 A and the voltage V=0.43 V. The equation for power P is P= I*V, so:
P=I*V=0.5*0.43=0.215 W
So the correct answer is that the power cell is supplying the motor with P=0.215 W of power.
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm
