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denis-greek [22]
2 years ago
5

8. A 2 kg flower pot weighing 20 N falls from a window ledge.

Physics
1 answer:
Alina [70]2 years ago
8 0

The force of the air resistance is 4 N.

The given parameters;

  • mass of the flower pot, m = 2 kg
  • weight of the flower pot, W = 20 N

Let the air resistance = F

Apply Newton's second law of motion to determine the force of the air resistance acting upward to oppose the motion of the pot falling downwards.

\Sigma F = ma\\\\W - F = ma\\\\a = \frac{W - F}{m} \\\\8 = \frac{20 - F}{2} \\\\20 - F = 16\\\\F = 20 - 16\\\\F = 4 \ N

Thus, the force of the air resistance is 4 N.

Learn more here: brainly.com/question/19887955

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A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light i
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Answer:

The appropriate solution is:

(a) \frac{1}{4}(I_o)

(b) \frac{1}{4} (u_o)

(c) \frac{1}{2}B_o

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

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On applying cross-multiplication, we get

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⇒     =I_o\times \frac{1}{4}

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(b)

As we know,

⇒ \frac{u}{u_o} =\frac{I}{I_o}

By putting the values, we get

⇒ u=\frac{1}{4}(u_o)

(c)

⇒ \frac{B^2}{B_o^2} =\frac{u}{u_o}

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4 0
3 years ago
Forces always act alone. true or false
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The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

(V_{f})^{2}=(V_{o})^{2}+2aS\\  So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}

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Now for distance

So

Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m

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