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amid [387]
3 years ago
15

The coefficient of friction on a surface is 0.25. A box requires 100N to slide it across the surface. What is the weight of the

box?
Physics
1 answer:
Novosadov [1.4K]3 years ago
5 0
So we know coefficient of f times normal force is friction. So do 100= .25 times x. Now solve for x. You get 400. So 400 is the normal force. And we know normal force equals weight in these types of problems so the answer is 400
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If its wavelength were doubled, its energy would be If its wavelength were doubled, its energy would be 4E. 2E. 12E. 14E.
ANEK [815]

The wavelength was doubled, and its energy will be increased by 4 times.

looking at the formula

energy E = MC^2

also, c = \lambda \times \nu

hence it is clear from above that energy is directly proportional to the square of the wavelength.

hence, The wavelength was doubled, and its energy will be increased by 4 times.

<h3>What is Wavelength?</h3>
  • The distance over which a periodic wave's shape repeats is known as the wavelength in physics.
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To learn more about wavelength with the given link

brainly.com/question/13533093

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1 year ago
The following are examples of physical properties except
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6 0
3 years ago
Read 2 more answers
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
Norma-Jean [14]

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

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7 0
2 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
dalvyx [7]

Answer:

Explanation:

a )  V = 3 cos(0.5t)

differentiating with respect to t

dv /dt = -3 x .5 sin0.5t

= -1.5 sin0.5t.

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when t = 3 s

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b )  dx/dt = 3 cos 0.5 t

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integrating on both sides

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At t = 2 s

x = 6 sin 1

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4 0
3 years ago
When the mass of the bottle is 0.125 kg, the average maximum height of the beanbag is m. When the mass of the bottle is 0.250 kg
Jet001 [13]

Answer:

when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.

when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.

when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.

when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.

Explanation:

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2 years ago
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