Answer:pelo o que eu sei é ..
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Explanation:
Answer:
20000 W
Explanation:
Power: This can be defined as the rate at which energy is dissipated or used. The S.I unit of power is Watt(W).
The expression of power is given as,
P = E/t.............................. Equation 1
Where P = power, E = Energy, t = time.
Given: E = 200 J, t = 0.01 s
Substitute into equation 1
P = 200/0.01
P = 20000 W.
Hence the average power = 20000 W
If Resistors are in series= The equivalent is the sum.
E.g R1 and R2 in series, R = R1 + R2.
If in Parallel, equivalent is Product/sum.
E.g If R1 and R2 in parallel, R = (R1*R2)/(R1+R2)
1.) 60 is parallel with 40 and both are then in series with 20.
60//40 = (60*40)/(60+40) = 2400/100 = 24
Now the 24 is in series with the 20
R = 24 + 20 = 44 ohms.
2.) 80 is in series with 40 and both are then in parallel with 40.
Solving the series, R = 80 + 40 =120.
Parallel: 120//40 = (120*40)/(120+40) = 4800/160 = 30
Equivalent Resistance = 30 ohms.
3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.
The 1st parallel = (100*100)/(100+100) = 10000/200 = 50
The 2nd parallel = (50*50)/(50+50) = 2500/100 = 25.
Solving the series = 50 + 25 =75 ohms.
Cheers.
Answer:
Explanation:
Let 100 m/s be the velocity of projection.
So horizontal component
= 100 cos42
= 74.31 m /s
Vertical component = - 100 sin 42 . in upward direction
66.91 m/s
Net displacement = 2.1 downwards ( + ve )
Using s = ut + 1/2 gt²
2.1 = - 66.91 t + .5 x 9.8 x t²
4.9 t² - 66.91 t - 2.1 = 0
t = 13.685 s
Horizontal distance covered
= 13.685 x 74.31
= 1016.93 m
If angle of projction is 40°
So horizontal component
= 100 cos40
= 76.60 m /s
Vertical component = - 100 sin 42 . in upward direction
64.27 m/s
Net displacement = 2.1 downwards ( + ve )
Using s = ut + 1/2 gt²
2.1 = -76.60 t + .5 x 9.8 x t²
4.9 t² - 76.60 t - 2.1 = 0
t = 15.659 s
Horizontal distance covered
= 15.659 x 76.60
= 1199.49 m
So horizontal range is increased , if angle of projection is increased .
I had the same question. I'm pretty sure it's 23 seconds. :)
