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andre [41]
2 years ago
11

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.16

kg and moves at v = 5.33 m/s. The circular path has a radius of R = 1.13 m. What is the minimum velocity so the string will not go slack as the ball moves around the circle?
Physics
2 answers:
ycow [4]2 years ago
4 0
We assign the variables: T as tension  and x the angle of the string
 The  <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx. 
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.

</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
kolezko [41]2 years ago
4 0

Answer:

v=3.57m/s

Explanation:

Hi, let's solve this!

The speed is minimal when the voltage T = 0N

This occurs when the balance is

T + m * g = m * v2 / r

Since the voltage is zero, then we cancel it and the equation is

m * g = m * v2 / r

We have the following data:

m = 0.16kg

g = 9.8m / s2

r = 1.13m

We cleared the speed and it is

v=\sqrt{g*r}

v=\sqrt{9.8m/s^{2}* 1.13m}

v=3.57m/s

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newton's second law states that when a net force acts on an object, it accelerates it.Explain how it would be possible for two o
Alika [10]

Answer:

See Explanation

Explanation:

According to Newton's second law of motion, the acceleration of a body is proportional to the net external force that acts on the body.

A body accelerated when it is acted upon by an unbalanced net external force.

When the external forces acting on a body are balanced, the effect of each force is cancelled by the other hence the body is not accelerated according to Newton's second law.

4 0
2 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Lemur [1.5K]

Answer:

a)   v = 0.9167 m / s,  b)  A = 0.350 m,  c)  v = 0.9167 m / s, d)  A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

          v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

           λ = x

           λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

          T / 2 = t

           T = 2t

period and frequency are related

           f = 1 / T

           f = 1 / 2t

we substitute

           v = x / 2t

           v = 5.50 / 2 3

           v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

          2A = y

           A = y / 2

           A = 0.700 / 2

           A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

      v = 0.9167 m / s

d) the amplitude is

           A = 0.500 / 2

           A = 0.250 m

4 0
2 years ago
A space probe lands on a newly discovered planet. A small canister is released from the probe and falls a distance of 3 m in 0.5
ivanzaharov [21]

Answer:

24m/s²

Explanation:

Given

Distance S = 3m

Time of fall = 0.5sec

Required

Acceleration due to gravity

Using the equation of motion

S = ut+1/2gt²

Substitute the given values

3 = 0+1/2g(0.5)²

3 = 1/2(0.25)g

3 = 0.125g

g = 3/0.125

g = 24

Hence the value for the acceleration of gravity on this new planet is 24m/s²

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notka56 [123]
The question just basically explained what happens
4 0
3 years ago
A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the
andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

F = mg

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m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

\sum \tau = 0

Regarding the forces we have,

3Mg-1mg=0

Re-arrange to find M,

M = \frac{m}{3}

M = \frac{50}{3}

M = 16.67Kg

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0
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