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andreev551 [17]
3 years ago
12

a 0.04kg ball tied to a string moves in a circle that has a radius of 0.70 m. If the ball is accelerating 43.2m/s, what is the t

angential velocity of the ball?​
Physics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

Tangential Velocity = 30.24 m/s

Explanation:

Given that,

Mass of ball, m = 0.04 Kg

Length of the string, r = 0.70 m

Acceleration of the ball, a = 43.2 m/s²

The tangential velocity of ball, V = ?

The centripetal force is given by the relation

        Fc = mV²/r  newton

where,   m - mass of body

              V - tangential velocity of body

               r - radius of the trajectory

Force applied on the ball to rotate on a circular path

                F = m x a newton

The applied force is equal to centripetal force.

So, equalizing the force equations

                     m x a = m V²/r

Therefore

                     V² = a x r

                      V = \sqrt{a X r}

Substituting the values

                       V = \sqrt{43.2 X 0.70}

                        V = 30.24 m/s

So, the tangential velocity of the ball is 30.24 m/s

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Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

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Now , we rearrange , and we get 6F=ma

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3 years ago
Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range
Maslowich

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

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8 0
3 years ago
A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?
Len [333]

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

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T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

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A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

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5 0
2 years ago
Read 2 more answers
A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,
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For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. <span>

An initial condition has been supplied by the problem: 

s=1.80m when t=0.245s 

<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2) 

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation): 

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2 

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<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

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0=(v)+(a)(t) 

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t=0.8714s 

<span>Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m<span>

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window: 

3.725m – 1.8m=1.925m</span>

 

Answer:

<span>1.925m</span>

3 0
3 years ago
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Paraphin [41]

Answer:

v = 3200 m/s

Explanation:

As we know that the frequency of the sound wave is given as

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so now we have

speed = wavelength \times frequency

so we will have

v = (8m) \times (400 Hz)

v = 3200 m/s

4 0
3 years ago
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