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Rudiy27
2 years ago
6

When a bicycle tire gets mud on it, the tire becomes slippery. What does the mud do to the tire? A. Increases the force of gravi

ty B. Increases the force of friction C. It decreases the force of friction D. It decreases the force of gravity
Physics
2 answers:
larisa86 [58]2 years ago
8 0
Question: <span>When a bicycle tire gets mud on it, the tire becomes slippery. What does the mud do to the tire?

C. It decreases the force of friction.

First mud is not a solid. It is more of a slow liquid like corn syrup. And when you are caught in mud or trying to ride past the mud you loose friction with the ground.
You could also resemble friction with traction. In a car. If there is water & oil mixed on the road and you go past it at a high mph range. You will loose traction and start skipping against the mixed substance. This reduces your traction (friction) with the ground.

Final Answer:

C. It decreases the force of friction
</span>
Flura [38]2 years ago
7 0
The answer to this is C.
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<h3>What is ohm's law in circuit?</h3>
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7 0
1 year ago
If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

<em>where</em>,

  • A = amplitude of the displacement of the wave.
  • k = wave number of the wave = \dfrac{2\pi }{\lambda}.
  • \lambda = wavelength of the wave.
  • x = horizontal displacement of the wave.
  • \omega = angular frequency of the wave = \rm 2\pi f.
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On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

therefore,

\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.

It is the required frequency of the wave.

3 0
2 years ago
A ball is at rest at the top of a hill until a boy kicked it with his foot. What is the force that causes motion in this scenari
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2 years ago
A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg
Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

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Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

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1120 = 20 v₁² + (10 − 5 v₁)²

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0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

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u₁ = 6 m/s, so v₁ = -3.78 m/s.  Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

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3 years ago
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