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Tasya [4]
2 years ago
14

What is the relationship between friction and velocity?

Physics
2 answers:
almond37 [142]2 years ago
8 0

\huge \sf \underline{ \purple{Åñwêr :}}

  • Friction is typically unrelated to speed. When the surfaces start to melt or otherwise change because of heat build up or inability to follow the surface then things will change.

  • Friction is related to velocity in that it acts to reduce velocity. In other words it causes moving objects to decelerate.

  • The friction force may also act as one of the forces in a moment that is causing motion. If the motion is rotary, or following any path, then the friction force may be simply a part of the system of forces that influences motion.

<h3>Hope It's Helps! :)</h3>

Umnica [9.8K]2 years ago
5 0
Friction is exerted when an object sliding over a surface and its formula
Ff =miu times normal force miu is the coefficient of friction
Velocity is how fast a body moves with in direction and velocity =displacement over time
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How are kinetic energy potential energy and thermal energy in a substance related?​
asambeis [7]

The energy associated with an object's motion is called kinetic energy. ... This is also called thermal energy – the greater the thermal energy, the greater the kinetic energy of atomic motion, and vice versa.

3 0
3 years ago
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Tems11 [23]

Answer:

The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.

Explanation:

5 0
3 years ago
Read 2 more answers
After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m&gt;s. Her 5.0-kg cat
Vinil7 [7]

To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_1= mass of the skier

m_2= mass of the cat

v_1 = initial velocity of skier

v_2 = initial velocity of cat

v_f= final velocity of both

Re-arrange to find V_f we have,

V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}

V_f = \frac{(60)(15)+(5)(-3.8)}{(60+5)}

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Once the final velocity is found it is possible to calculate the change in kinetic energy, so

\Delta KE = KE_i-KE_f

\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2

\Delta KE = \frac{1}{2}((60)(15)^2+(5)(-3.8)^2)-\frac{1}{2}(60+5)(13.55)^2

\Delta KE = 819.1J

Therefore the amount of kinetic energy converted in to internal energy is 819J

3 0
3 years ago
Pls help :3 ♥️ I rly need this turned in
Lerok [7]

Answer:

It does, it takes 50ml

Explanation:

5 0
3 years ago
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