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Ilia_Sergeevich [38]
3 years ago
9

What is friction and its types? ​

Physics
2 answers:
Kitty [74]3 years ago
3 0
Friction is defined as the force that opposes the motion of a solid object over another. There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction.
jeka57 [31]3 years ago
3 0

Answer:

Friction is defined as the force that opposes the motion of a solid object over another. There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. Friction and normal force are directly proportional to the contacting surfaces and it doesn’t depend on the hardness of the contacting surface. With the increase in relative speeds, the sliding friction reduces whereas fluid friction increases with the increase in the relative speed, also fluid friction is dependent on the viscosity of the fluid.

In sliding motion, each point on the body has only translational or linear motion. But in case of rolling motion, different points have a combination of linear as well as rotational motion.

Types Of Friction

Static Friction

Sliding Friction

Rolling Friction

Fluid Friction

Static Friction

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A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng
frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

Q_{1} =\frac{W}{n}

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

Q_{1} =\frac{750}{0.35} =2142.86kJ

Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ

c) The efficiency of Carnot engine is:

n=1-\frac{300K}{550K} =0.455

The heat transfer to the heat engine is:

Q_{1c} =\frac{750}{0.455} =1648.35kJ

The heat transfer from the heat engine is:

Q_{2c} =1648.35-750=898.35kJ

4 0
3 years ago
The scanner records the time from when a ultrasound wave is emitted to when its reflection is received
telo118 [61]

Explanation:

The scanner records the time from when a ultrasound wave is emitted to when its reflection is received. A technician calculates the depth of the reflection using the equation as :

\text{depth}=\dfrac{1}{2}\times \text{speed of ultrasound in patient}\times \text{time recorded by scanner}

The distance covered by ultrasonic wave when it was emitted and gets reflected is 2d. Speed is given by :

v=\dfrac{d}{t}

d is distance and t is time

Here, d = 2d

So, the factor of (1/2) is used because the distance covered by the wave is 2 times when it was emitted and received.

3 0
3 years ago
two forces equal in magnitude and opposite in direction act at the same point on an object. is it possible for there to be a net
elena55 [62]

If the forces are equal, at a distance equidistant it is not possible to act a pair on the body since both torques cancel each other. Being of the same magnitude and in the opposite direction, the sum of the torques will be zero.

5 0
3 years ago
Two cars are traveling at the same constant speed v. Car A is moving along a straight section of the road, while B is rounding a
Kamila [148]

Answer:

Car A is not accelerating, but car B is accelerating

Explanation:

Car A is not accelerating because , it is moving with uniform speed in a particular direction . Therefore its velocity too is not changing .

But car B is accelerating because , though its speed is uniform but its velocity is changing due to change in direction . On a circular path , direction of speed changes every moment . Therefore it has an acceleration called centripetal acceleration . Hence car B has acceleration.

7 0
4 years ago
Two cars are traveling around identical circular racetracks. Car A travels at a constant speed of 20 m/s. Car B starts at rest a
Tomtit [17]

Answer:

b. it has the same centripetal acceleration as car A.

Explanation:

According to the question, the data provided is as follows

Constant speed of car A = 20 m/s

Constant tangential acceleration until its speed is 40 m/s

Based on the above information, the true statement is the same centripetal acceleration as car A because

As we know that

Centripetal acceleration is

= \frac{V^2}{r}

where,

V^2 = velocity

r = radius of the path

Now if both car A and car B moving in the same or identical circular path having the same velocity so in this case there is the same centripetal acceleration for that particular time

hence, the second option is correct

3 0
4 years ago
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