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arlik [135]
3 years ago
8

PLEASE help me with these questions as soon as possible i need to turn it in!!!!

Physics
1 answer:
Sloan [31]3 years ago
3 0
A. I believe this might be the answer but
6 J/60 J= .1
.1 x 100= 10%

b. I'm not sure of this question


19. I'm not sure if I did this correct way or if there is a better way (I'm rusty) 
300 J/.50 %= 600
600/100= 6 (answer)


20. (200 J)(.75%) = 150 (answer)
Check:
Efficiency: output/input x 100
150 J/200 J= .75
.75 x 100= 75%
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The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?
anastassius [24]
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
\rho =  \frac{AR}{L}
where
\rho is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
A=\pi r^2
where r is the radius of the wire. Substituting in the previous equation ,we find
\rho =  \frac{\pi r^2 R}{L}

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
\rho' =  \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4  \frac{\pi r^2 R}{L}   = 4 \rho
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5 0
3 years ago
Read 2 more answers
A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.
matrenka [14]

Answer:

Part a)

T = 42 N

Part b)

v_f = 11.8 m/s

Part c)

t = 1.7 s

Part d)

F = 159.7 N

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

mg - T = ma

for uniform cylinder we will have

TR = I\alpha

so we have

T = \frac{1}{2}MR^2(\frac{a}{R^2})

T = \frac{1}{2}Ma

now we have

mg = (\frac{M}{2} + m)a

a = \frac{mg}{(\frac{M}{2} + m)}

a = \frac{15 \times 9.81}{(6 + 15)}

a = 7 m/s^2

now we have

T = \frac{12 \times 7}{2}

T = 42 N

Part b)

speed of the bucket can be found using kinematics

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(7)(10)

v_f = 11.8 m/s

Part c)

now in order to find the time of fall we can use another equation

v_f - v_i = at

11.8 - 0 = 7 t

t = 1.7 s

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

F = T + Mg

F = 42 + (12 \times 9.81)

F = 159.7 N

5 0
2 years ago
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