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garri49 [273]
2 years ago
12

A gymnast of mass 70.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

hat the rope does not stretch. Use the value 9.81m/s2 for the acceleration of gravity. Calculate the tension T in the rope if the gymnast hangs motionless on the rope.
Physics
1 answer:
DiKsa [7]2 years ago
3 0

Answer:

T = 686.7N

Explanation:

For this exercise we will use Newton's second law in this case there is no acceleration,

      ∑ F = ma

      T -W = 0

The gymnast's weight is

     W = mg

We clear and calculate the tension

     T = mg

     T = 70 9.81

     T = 686.7N

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A man can row at 6kmhr in still water and want to cross a river to a position exactly opposite his starting point. If the river
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Answer:

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2 years ago
A large ebony wood log, totally submerged, is rapidly floating down a flooded river. If the mass of the log is 165 kg, what is t
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Answer:

F = 1618.65[N]

Explanation:

To solve this problem we use the following equation that relates the mass, density and volume of the body to the floating force.

We know that the density of wood is equal to 750 [kg/m^3]

density = m / V

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m = mass = 165[kg]

V = volume [m^3]

V = m / density

V = 165 / 750

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The floating force is equal to:

F = density * g * V

F = 750*9.81*0.22

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2 years ago
A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is
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4 0
2 years ago
Read 2 more answers
A yo-yo is made from two uniform disks, each with mass m and radius R, connected by a light axle of radius b. A light, thin stri
schepotkina [342]

Answer:

linear acceleration

a = \frac{2g}{2 + \frac{R}{b}}

angular acceleration

\alpha = \frac{2g}{R(2 + \frac{R}{b})}

Explanation:

As we know that the force due to tension force is upwards while weight of the disc is downwards

so we will have

2mg - T = 2ma

also we have

Tb = (\frac{1}{2}mR^2 + \frac{1}{2}mR^2)\alpha

now we have

Tb = mR^2(\frac{a}{R})

T = \frac{mRa}{b}

now we have

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a(2 + \frac{R}{b}) = 2g

so we have

linear acceleration

a = \frac{2g}{2 + \frac{R}{b}}

angular acceleration

\alpha = \frac{2g}{R(2 + \frac{R}{b})}

4 0
2 years ago
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