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3 years ago

12

If the work required to speed up a car from 11 km/h to 21 km/h is 6.0×103 J , what would be the work required to increase the ca

r’s speed from 21 km/h to 33 km/h

1 answer:

goblinko [34]3 years ago

6 0

Explanation:

We need convert the velocities first to m/s and we get the following:

v2 = 21 km/hr = 5.8 m/s

v1 = 11 km/hr = 3.1 m/s

We need to find the mass of the car also for later use do using the work-energy theorem:

$delta \: w = \frac{1}{2} m(v \frac{2}{2} - v\frac{2}{1} )$

6.0x10^3 J = (0.5) m [(5.8)^2 - (3.1)^2]

or

m = 499.4 kg

Now we determine work needed delta W to change its velocity from 21 km/hr to 33 km/hr

v2 = 33 km/hr = 9.2 m/s

v1 = 21 km/hr = 5.8 m/s

delta W = (0.5)(499.4)[(9.2)^2 - (5.8)^2]

= 1.3 x 10^4 J

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Read 2 more answers

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sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

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mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

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radius of the final orbit, $r=0.5\ m$

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$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

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$v=1.2048\times 10^{7}\ m.s^{-1}$

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<u>We firstly find the relativistic mass as:</u>

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

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$KE=m'.c-m.c$

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Answer:

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b) 294000 N

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(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

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Answer:

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