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3 years ago

A force of 120 N is applied to the front of a sled so as to pull the sled a distance of 165

Physics

1 answer:

Marrrta [24]3 years ago

8 0

Answer:

<h3>The answer is 19,800 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

force = 120 N

distance = 165 m

We have

work done = 120 × 165

We have the final answer as

Hope this helps you

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A solid sphere of radius 30cm is uniformly charged to 100nC. a) What is the volume charge density of the sphere? b) What is the

g100num [7]

Answer

given,

total charge Q = 100 n C

= 100 × 10⁻⁹ C

radius of the solid sphere = 30 cm

= 0.3 m

Volume of sphere = $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi\times 0.3^3$

=0.113 m³

a) volume charge density

$\rho = \dfrac{10^{-7}}{0.133}$

ρ = 8.85 × 10⁻⁷ C/m³

b) at r = 10 cm = 0.1 m

charge in the sphere at radius

$Q = \dfrac{4}{3}\pi\times 0.1^3\time \rho$

= 3.7037 \times 10^{-9}C[/tex]

Field strength

$E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}$

$E_1 = \dfrac{3.7037 \times 10^{-9}}{4\pi \times 8.85\times 10^{-12}\times 0.1^2}$

= $3.33 \times 10^3 N/C$

at r = 20 cm = 0.2 m

$Q = \dfrac{4}{3}\pi\times r^3\time \rho$

$E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}$

$E_1 = \dfrac{ \dfrac{4}{3}\pi\times r^3\time \rho}{4\pi \epsilon_0 r^2}$

$E_1 = \dfrac{\rho}{3 \epsilon_0}$

$E_1 = \dfrac{0.2\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}$

= $6.66 \times 10^3 N/C$

at r = 30 cm

$E_1 = \dfrac{\rho}{3 \epsilon_0}$

$E_1 = \dfrac{0.3\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}$

= 9.99 N/C

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