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Elena L [17]
2 years ago
5

A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the chair is i

n motion, a 327 N horizontal force keeps it moving at a constant velocity.
a)Find the coefficient of static friction between the chair and the floor.
b)Find the coefficient of kinetic friction between the chair and the floor.
Physics
1 answer:
weqwewe [10]2 years ago
4 0

Answer:

Here's what you know, m=25 kg, Ff=365 N, g=9.8 m/s^2, use the formula Fg = mg, Fg =25*9.8, Fg =245 N, then use the formula Ff <= ตsFn, 365=245ตs, solve for ตs,  ตs=1.5.

Here's what ya know, m=25 kg, Ff=327 N, the normal force of gravity will be the same as in part A.  Now use the formula Ff = µkFn, 327=245 µk, solve for µk, µk=1.3.

Explanation:

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Why football players boots have spikes their sole<br>Give short and sweet answer​
Marizza181 [45]

Answer:

it helps with balance and speed.

"The football shoes have spikes or studs because the studs or spikes provides larger frictional force than normal shoes while running on the grass. The studs prevents player from slipping on the grass and help to run faster and change direction quickly without slipping"

3 0
3 years ago
Read 2 more answers
A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the
Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

qvB = m\frac{v^2}{r}

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

T=\frac{2\pi r}{v}

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0
3 years ago
Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an
tankabanditka [31]

Answer:4-strikes the plane at same time as the other body

Explanation:

Given

If both bodies is falling on a horizontal plane and second body is given an acceleration in horizontal direction then it does not change the time to reach the Horizontal Plate as there is no change in vertical direction.

Horizontal acceleration will give only horizontal range and horizontal velocity.

8 0
3 years ago
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8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of ball.

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of bottle.

We are given:

m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s

Putting values in above equation, we get:

(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0
3 years ago
in a cricket match there are 5000 spectators counted 10 by 10 the number of significant figure in the measurement will be
BabaBlast [244]

Answer:

\boxed{3 \ Significant \ figures}

Explanation:

Total spectators = 5000

Counted by the groups of ten, So at last the result will be:

=> 5000/10 = 500

Significant figures in 500 are 3

8 0
3 years ago
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