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3 years ago

The amount of friction divided by the weight of an object forms a unit less number called the

Physics

1 answer:

Romashka [77]3 years ago

3 0

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

$F=\mu N$

N is normal force.

$\mu$ = coefficient of friction

$\mu=\dfrac{F}{N}$

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The answer should be 88.56

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3 years ago

Two identical balls move directly toward each other with equal speeds. how will the balls move if they collide and stick togethe

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3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

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3 years ago

At what point will the electric field of a charged object be strongest?

puteri [66]

Answer:

The answer is C because they have to be close to be able to interact

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2 years ago

A gas has an initial volume of 2.5 L at a temperature of 275 K and a pressure of 2.1 atm. The pressure of the gas increases to 2

olchik [2.2K]

Answer:

2.10L

Explanation:

Given data

V1= 2.5L

T1= 275K

P1= 2.1atm

P2= 2.7 atm

T2= 298K

V2= ???

Let us apply the gas equation

P1V1/T1= P2V2/T2

substitute into the expression we have

2.1*2.5/275= 2.7*V2/298

5.25/275= 2.7*V2/298

Cross multiply

275*2.7V2= 298*5.25

742.5V2= 1564.5

V2= 1564.5/742.5

V2= 2.10L

Hence the final volume is 2.10L

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3 years ago