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3 years ago

If a 15 N box is lifted a distance of 3 m, how much work is done?

Physics

2 answers:

Naily [24]3 years ago

7 0

Answer:

W=45J

Explanation:

W=Fd

W=15(3)=45

W=45J

Oxana [17]3 years ago

5 0

Answer:

work done = 45 j

Explanation:

work done ( W.D ) = force × displacement

=》W.D = 15 N × 3 m = 45 joule

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Read 2 more answers

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

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vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

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Answer:

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