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jasenka [17]
3 years ago
13

A 1.35-kg ball is tied to a 1.10-m long string is being spun in a vertical circle at a constant speed and with a period of 2.10

s. What is the maximum tension in the string?
Physics
2 answers:
ikadub [295]3 years ago
6 0

Answer:

T = 13.3 N

Explanation:

In this exercise we use Newton's second law, in the lower birth of the circle the tension and force are opposite, so the tension has its maximum value

          T - W = m a

The acceleration is centripetal

         a = v² / r = w² r

We replace

               T = mg + m w² r

The angular velocity is related to the period

              w = 2π f = 2π / T₀

             T₀ = 2.10 s

             T = m (g + 4π² r / T₀²)

Let's calculate

             T = 1.35 (9.8 + 4π² 1.10 /2.10²)

             T = 13.3 N

777dan777 [17]3 years ago
6 0

Answer:

26.51 N

Explanation:

Using

F = mω²r + mg ....................... Equation 1

Where F = maximum tension in the string, ω = angular velocity, r = radius of the circle formed/ Length of the ball, g = acceleration due to gravity

Also,

ω = 2π/T........................ Equation 2

Where T = Period, π = Pie

Substitute equation 2 into equation 1

F = [mr(2π/T)²]+mg

F = [mr4π²/T²]+mg.................... Equation 3

Given: m = 1.35 kg, r = 1.10 m, T = 2.10 s, g = 9.8 m/s², π = 3.14

Substitute into equation 3

F = 1.35(1.1)(4)(3.14²)/2.1² + 1.35(9.8)

F = 13.28+13.23

F = 26.51 N.

Hence the maximum tension in the string = 26.51 N

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Answer

given,

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jenyasd209 [6]

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A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
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Part a)

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Part b)

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Explanation:

Part a)

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here we know that

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now we have

75 = 10 Log(\frac{I}{10^{-12}})

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Part b)

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here we know that

A = \frac{P_o}{Bk}

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I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

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Answer:

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A coaster sliding against a table.

Gravity:-

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The force that causes a car to coast downhill even when you aren't stepping on the gas.

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