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3 years ago

A 25 n object requires a 5.0 n to start moving over a horizontal surface. what is the coefficient of static friction?

Physics

2 answers:

Cerrena [4.2K]3 years ago

7 0

Answer:

0,2 would be the coefficient of friction.

Explanation:

Remember that the force necessary to move an object over a horizontal surface is the weight of the object multiplied by the coeficient of friction, an the formula is this:

$Friction= u*N$

Where u is the coefficient of static friction and N is the normal force, or in this case the weight of the object so if we solve for U the formula would be like this:

$u=\frac{Friction}{N}$

Now we just insert the data we know:

$u=\frac{Friction}{N}\\u=\frac{5N}{25N}\\ u=0,2$

So the coefficient of friction would be 0.2

olganol [36]3 years ago

3 0

Static friction is the friction that exists between two or more solids that are not moving with a relative speed. To calculate the static friction coefficient we use the formula Fs=us × n where Fs is the static friction , us is the coefficient of static friction and the n is the normal force.
thus the coefficient of static friction will be 5 N÷ 25 N = 0.2
Hence 0.2 is the coefficient of static friction

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In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

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3 years ago

A 3.3 kg ball sits on the ground and is kicked with a FAPP of 36N

jeyben [28]

a) 32.3 N

The force of gravity (also called weight) on an object is given by

W = mg

where

m is the mass of the object

g is the acceleration of gravity

For the ball in the problem,

m = 3.3 kg

g = 9.8 m/s^2

Substituting, we find the force of gravity on the ball:

$W=(3.3)(9.8)=32.3 N$

b) 48.3 N

The force applied

$F_{app} = 36 N$

The ball is kicked with this force, so we can assume that the kick is horizontal.

This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:

$F=\sqrt{W^2+F_{app}^2}$

And substituting

W = 32.3 N

Fapp = 36 N

We find

$F=\sqrt{32.3^2+36^2}=48.3 N$

c) $14.6 m/s^2$

The ball's acceleration can be found by using Newton's second law, which states that

F = ma

where

F is the net force on an object

m is its mass

a is its acceleration

For the ball in this problem,

m = 3.3 kg

F = 48.3 N

Solving the equation for a, we find

$a=\frac{F}{m}=\frac{48.3}{3.3}=14.6 m/s^2$

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ale4655 [162]

Answer:

TRUE

Explanation:

The answer is true.

Balance forces acting on a body will not change the motion of the body because the body experiences no net resultant force in one direction. When any body experiences equal forces with opposite directions, the net force or the resultant force experience by the body is zero.

In case of an unbalanced forces, there is a net force acting in one direction and so it causes the body to change in its state of motion in the direction of the net force.

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Answer:

Explanation:

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