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3 years ago

Ayuda porfavorrrrrrrrrrr

Physics

1 answer:

zhuklara [117]3 years ago

8 0

what do you need help with can i help with something or do you what me to give you the answer ask me anething and i got youExplanation:

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A bowling ball of mass and radius is launched down the alley with speed and no rotation. Kinetic friction with the floor, , slow

lianna [129]

Answer:

Explanation:

Given that,

Mass of ball =M

Radius of ball is =R

Coefficient of kinetic friction =u

Initial speed of ball =Vo

The weight of the body acting downward is

W=Mg.

The normal reaction is acting upward and it is given as

From Newton law

N=W=mg

Frictional force is given as

Fr=µN

Fr= µmg

This is the only force on the x-axis

Then,

ΣF = ma

-Fr=ma

-µmg=ma

Divide through by m

a= -µg

The negative show that it is decelerating

So using equation of motion

V=u+at

Where V is final velocity,?

u is initial velocity =Vo

And a is acceleration =-µg

Then, the velocity at any point in time is given as

V = Vo - µgt

The angular acceleration that sets the ball rotating with increasing angular velocity in anticlockwise direction whose magnitude ω, at any instant t, is given by

ω = αt.

Also, V=ωR

V=αtR

To get angular acceleration

Further, the only force that produces a torque about the centre is fk. This torque is of magnitude fkR, acting in anticlockwise direction producing an anticlockwise angular acceleration, α, of the ball about its center given as

fk•R = Icm •α,

Icm for a sphere is 2/5MR²

µMg•R = (2/5)MR² α

Divide both side by MR

µg= (2/5)Rα

α = 5µg/2R.

From above

V = Vo - µgt, then, V=αtR

αtR= Vo - µgt

αtR + µgt= Vo

t(αR + µg)=Vo

t=Vo/(αR + µg)

Since α = 5µg/2R

t=Vo/( 5µg/2R • R + µg)

t = Vo/( 5µg/2+ µg)

t= Vo/(7µg/2)

t=2Vo/7µg

So, the time is given as

t = 2Vo/7µg

And the velocity at any time is given as

V = Vo - µgt,

5 0

3 years ago

An ideal gas initially at 4.00atm and 350 K is permitted

Nuetrik [128]

Explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is $P_{2}$ and final temperature is $T_{2}$ .