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3 years ago

/How much kinetic energy does an 80 kg man have while running at 1.5 m/s

Physics

1 answer:

igomit [66]3 years ago

7 0

K.E = 0.5 * m * v^2
= 0.5 * 80 * (1.5)^2
K.E = 90 joule

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Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

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Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

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Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

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We need to calculate the magnitude of the electric field at r = 11.5 cm

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$E=\dfrac{kq}{r^2}$

Where, q = charge

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r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

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3 years ago

How do I do this problem?

Aleks04 [339]

Use Charles Law: V1/T1 = V2/T2

0.30 m^3/27 C = V2/127 C

27V2 = 127 * 0.3

V2= 38.1/27 = 1.4 m^3

5 0

3 years ago