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1 year ago

A 850-kg sports car accelerates from rest to 95 km/h in 6.8 s .

Physics

1 answer:

swat321 year ago

5 0

From the calculations, the power expended is 43650 W.

<h3>What is the power expended?</h3>

Now we can find the acceleration from;

v = u + at

u = 0 m/s

v = 95 km/h or 26.4 m/s

t = 6.8 s

a = ?

Now

v = at

a = v/t

a = 26.4 m/s/ 6.8 s

a = 3.88 m/s^2

Force = ma = 850-kg * 3.88 m/s^2 = 3298 N

The distance covered is obtained from;

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (26.4)^2/2 * 3.88

s = 696.96/7.76

s = 90 m

Now;

Work = Fs

Work = 3298 N * 90 m = 296820 J

Power = 296820 J/ 6.8 s

= 43650 W

Learn more about power expended:brainly.com/question/11579192

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A student team is to design a human powered submarine for a design competition. The overall length of the prototype submarine is

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Answer:

a) The speed is 61.42 m/s

b) The drag force is 10.32 N

Explanation:

a) The Reynold´s number for the model and prototype is:

$Re_{m} =\frac{p_{m}V_{m}L_{m} }{u_{m} }$

$Re_{p} =\frac{p_{p}V_{p}L_{p} }{u_{p} }$

Equaling both Reynold's number:

$\frac{p_{p}V_{p}L_{p} }{u_{p} }=\frac{p_{m}V_{m}L_{m} }{u_{m} }$

Clearing Vm:

$V_{m} =\frac{p_{p}V_{p}L_{p} u_{m} }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s$

b) The drag force is:

$\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2} } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2} } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2} } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2} } =10.32N$