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scoray [572]
2 years ago
14

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

Physics
1 answer:
swat322 years 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

#SPJ1

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Answer:

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Explanation:

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7 0
3 years ago
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Ivahew [28]

Answer:

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Explanation:

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lakkis [162]
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7 0
3 years ago
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elements found in the same ______ have similar chemical properties. What should go in the blank? Row or column?
Temka [501]
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3 0
4 years ago
A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is
neonofarm [45]

Answer:

a) 25.5°(south of east)

b) 119 s

c) 238 m

Explanation:

solution:

we have river speed v_{r}=2 m/s

velocity of motorboat relative to water is v_{m/r}=4.2 m/s

so speed will be:

a) v_{m}=v_{r}+v_{m/r}

solving graphically

v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}

     =4.7 m/s

Ф=tan^{-1} (\frac{v_{r}}{v_{m/r}} )

  =25.5°(south of east)

b) time to cross the river: t=\frac{w}{v_{m/r}}=\frac{500}{4.2}=119 s

c) d=v_{r}t=(2)(119)=238 m

note :

pic is attached

6 0
4 years ago
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