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sergeinik [125]
3 years ago
8

A man pushed the box with a force of 20 N. The mass of the box doubled, but he still pushed with a force of 20 N.

Physics
1 answer:
Nuetrik [128]3 years ago
7 0

Answer:

Option (c) Acceleration is 1/2x the original acceleration

Explanation:

To know the the correct answer to the question, we shall determine the acceleration of the box in each case.

Case 1:

Mass (m) = m

Force (F) = 20 N

Acceleration 1 (a₁) =?

F = ma₁

20 = m × a₁

Divide both side by m

a₁ = 20 / m

Case 2:

Mass (m) = 2m

Force (F) = 20 N

Acceleration 2 (a₂) =?

F = ma₂

20 = 2m × a₂

Divide both side by 2m

a₂ = 20 / 2m

a₂ = 10 / m

Finally, we shall determine the acceleration of the box after the mass was doubled. This can be obtained as illustrated below:

Acceleration 1 (a₁) = 20 / m

Acceleration 2 (a₂) = 10 / m

a₂ : a₁ = 10 / m : 20 / m

a₂ / a₁ = 10 / m ÷ 20 / m

a₂ / a₁ = (10 / m) × (m / 20)

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

Thus, the acceleration of the box after the mass was doubled is ½ times the original acceleration.

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The peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

<h3>Relationship between electric and magnetic field</h3>

The relationship between electric and magnetic field at a given peak electric field is given as;

c = (E₀) / (B₀)

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B₀ = E₀ / c

B₀ = (2.9) / (3 x 10⁹)

B₀ = 9.67 x 10⁻¹⁰ T

Thus, the peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

Learn more about peak magnetic field here: brainly.com/question/24487261

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2 years ago
If an engine does 660 J of work in 10 seconds, its average power is ...
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Anyone know which wire matches the other one? for all 4. Need help! Thanks :-)
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Which winds are affected by specific landforms on earth's surface?
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A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo
Serggg [28]

Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

where we take g=9.80\,\frac{\rm m}{\mathrm s^2}.

a. The maximum altitude occurs at the point during which v_y=0:

159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

At this point, the rocket has an altitude of

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

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3 years ago
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