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Blababa [14]
3 years ago
14

Formular for coefficient of friction​

Physics
1 answer:
Bumek [7]3 years ago
7 0

Answer:

hope you like it mark as brainliest

Explanation:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact.

The coefficient of friction (fr) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together. It is represented by the equation: fr = Fr/N.

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A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg
Naddik [55]

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

4 0
3 years ago
A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc
Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

<h3>Fբ = 75 N</h3>

  • Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

<h3>Fₙ = 25 N</h3>

  • Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

<h3>a = 0.5 m/s²</h3>

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0
2 years ago
Jeremiah is conducting an investigation about the water cycle. He is given the following
Vlada [557]

The water cycle outlines the continuous water movement in liquid, solid and gaseous state between locations on the Earth's surface.

  • The glass jar represents the lake while the atmosphere is represented by the space above the water, and the sky is represented by the (clear) plastic wrap

Arrangement description and Processes;

The processes of the water cycle includes;

  • Evaporation;
  • Condensation
  • Precipitation
  • Sublimation
  • Runoff
  • Infiltration

The arrangement of the materials is as follows;

  • Place the glass jar (the lake) containing water and the lamp (the Sun) side by side, such that the lamp light shines on the water surface

  • Cover the glass jar by wrapping the plastic wrap (the sky) around it to prevent the escape of water vapor when the water is hot.

  • Switch on the lamp so that it heats the water by radiation heat transfer

Observed processes;

The processes demonstrated by the above experiment includes;

1) Evaporation: As the water in the glass jar becomes warmer, the level of the water in the jar can be observed to decrease slightly due to evaporation

2) Condensation: Fog formation, Clouds

When hotter, the water surface as seen through the clear plastic wrap becomes less clearer due to evaporation, and condensation of the vapor while floating above the water surface, similar to the clouds seen in the sky.

3. Precipitation: Rain;

The clear plastic wrap covering the top of the glass jar, prevents the movement of the vapor further away, such that the tiny condensed vapor gather together, to form big droplets under the plastic wrap that falls back into the jar, which is similar to the process of rainfall

The above processes are repeated as more water evaporates from the jar condenses on the plastic wrap and falls back into the jar, showing the process by which water is recycled from the lake into the atmosphere and back to the lake.

Learn more here:

brainly.com/question/2430469

4 0
2 years ago
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
Which statement is true for a sound wave entering an area of warmer air?
Sergio [31]
The answer is A........The sound is transferred by collisions of molecules. Therefore sound waves will travel faster on warm air because collisions of molecules of air in warm air are greater.
3 0
3 years ago
Read 2 more answers
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