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Zolol [24]
3 years ago
6

Different masses are hung on a spring scale calibrated in newtons. The force exerted by gravity on 1.0 kg is shown in the image

below in newtons.
An image of a spring scale with a 1 kilogram weight attached. The scale reads 9.8 N.

What would be the force exerted by gravity on 6.5 kg?

6.5 N
9.8 N
64 N
98 N
Physics
2 answers:
Lunna [17]3 years ago
6 0
6.5 x 9.8
=63.7
so the answer is 64N
Arada [10]3 years ago
3 0

Answer:

F = 64 N

Explanation:

When 1 kg of weight attached with the spring, the scale reads 9.8 N. Weight of an object is the product of mass and acceleration due to gravity. Let F is the force exerted by gravity on 6.5 kg. It can be calculated as :

F=mg

F=6.5\ kg\times 9.8\ m/s^2

F = 63.7 N

or

F = 64 N

So, the force exerted by gravity on 6.5 kg is 64 N.

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Answer:

σ = 3.402 KPa ,  γ = 0.25 , G = 13.608 KPa

Explanation:

Given:-

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- The displacement of upper surface, x = 10 mm

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Find the shearing stress, shearing strain and  shear modulus.​

Solution:-

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- The shear effect results in a stress in the gelatin block.

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                        σ = F / A

Where,

           A : The surface area of the object that experiences the shear force.

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          A = ( 0.120 )*( 0.120 ) = 0.0144 m^2

Therefore,

                     σ = 49 / 0.0144

                     σ = 3.402 KPa

- The shear strain ( γ ) is the measurement of change in dimension per unit depth of the block.

- The top surface undergoes a displacement of ( x ). The height of the top surface of the gelatin block is L = 40 mm.

Hence,

                    γ = x / L

                    γ = 10 / 40

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- It is mathematically expressed as a ratio of shear stress  ( σ ) and shear strain ( γ ):

                   G =  σ / γ

                   G = 3.402 / 0.25

                   G = 13.608 KPa

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