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kenny6666 [7]
3 years ago
8

A. A PH202 student lives next to a construction site and sees a crane with a wrecking ball demolish the building next door. The

wrecking ball swings along the wall between her house and the neighbor’s house. In an effort to determine the length of the cable on the wrecking ball the student builds a pendulum using a random rock and a string. Her pendulum turns out to be 0.500m long. While she plays with her pendulum she realizes that the wrecking ball swings back and forth in the same amount of time that it takes the rock to complete 5 full oscillations. What is the length of the cable on the wrecking ball?
Physics
1 answer:
netineya [11]3 years ago
8 0

Answer:

The length of cable is 12.5 m

Explanation:

Since, the wrecking ball completes 1 oscillation, in the same time, as it takes for the rock to complete 5 oscillations.

Therefore,

Time Period of Wrecking Ball = 5 (Time Period of Rock)

Since,

Time Period of  Pendulum = 2π√(L/g)

Therefore,

2π√(L₁/g) = 5[2π√(L₂/g)]

√L₁ = 5√L₂

Squaring on both sides:

L₁ = 25 L₂

where,

L₁ = Length of Cable = ?

L₂ = Length of string = 0.5 m

Therefore,

L₁ = 25 (0.5 m)

<u>L₁ = 12.5 m</u>

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Answer:

Yes, the velocity would also be zero.

Explanation:

Acceleration is the change in velocity over time, therefore, there has to be a change in velocity for something to accelerate. which means without acceleration, the object has no velocity.

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3 years ago
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Rus_ich [418]

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Explanation:

7 0
3 years ago
food product with 10 kg mass is being transported to thesurface of the moon,where the acceleration due to gravity is1.624 m/s2;
larisa [96]

Answer:

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Explanation:

g = Acceleration due to gravity

m = Mass = 10 kg

Weight on Earth

W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N

Converting to lbf

98.1\times 0.22481=22.053861\ lbf

On Moon

W=10\times 1.624\\\Rightarrow W=16.24\ N

Converting to lbf

16.24\times 0.22481=3.6509144\ lbf

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

5 0
3 years ago
A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
Kisachek [45]

Answer:

The force is  F= 46.25kN

Explanation:

The diagram for this question is shown on the first uploaded image  

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

 v_2 is the is the speed of water at the nozzle which can be mathematically evaluated as

                      v_2 = \frac{R}{A_n}

substituting  0.5m^3/s for R and \frac{\pi}{4}(12*\frac{1m}{100} )^2 for A_n

                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

\rho is he density of water with value \rho =1000 kg /m^3

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

                                 = 21.99kN

At Equilibrium the summation of the of force on the horizontal axis is zero

                  i.e   \sum F_x =0

=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

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 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

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                      F = \sqrt{F_x^2 + F_y^2}

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6 0
3 years ago
Read 2 more answers
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Digiron [165]

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<em>V = 24.25 cos(377 t)</em>

V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.

6 0
3 years ago
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