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irakobra [83]
3 years ago
15

Which of the following best defines spring constant ?

Physics
1 answer:
katovenus [111]3 years ago
7 0

Answer:

A

Explanation:

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A block of gelatin is 120 mm by 120 mm by 40 mm when unstressed.
ycow [4]

Answer:

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

Explanation:

Given:-

- The dimension of gelatin block = ( 120 x 120 x 40 ) mm

- The applied force, F = 49 N

- The displacement of upper surface, x = 10 mm

Find:-

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

Solution:-

- The shear stress is the internal pressure created in an object opposing the applied action ( Force, moment, bending, or torque ).

- A force of F = 49 N was applied parallel to the top surface of the gelatin block.

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

- The formulation of stress ( σ ) is given below:

                        σ = F / A

Where,

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

- The top surface have the following dimensions:

          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

                    γ = 0.25

- The shear modulus or the modulus of rigidity ( G ) is a material intrinsic property that signifies the amount of resistive stress to any cause of deformation.

- 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

7 0
3 years ago
a stone is projected vertically up from the top of a tower 73.5m with velocity 24.5 m/s . find the time taken by the stone to re
hoa [83]

The stone's altitude at time t is given by

y=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\dfrac{\rm m}{\mathrm s^2} is the acceleration due to gravity. The stone reaches the ground when y=0:

0=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=7.11\,\rm s

6 0
3 years ago
For a solar eclipse to occur which of the following alignments must be necessary
katovenus [111]
Very specific alignment of the Sun, Earth, and Moon. If the Moon is lined up precisely with the Sun from the Earth's point of view, the Moon will block Sunlight from reaching the Earth, causing a solar eclipse.
6 0
3 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
GuDViN [60]

Answer:

7.2\cdot 10^{-19} J

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

\Delta U = q \Delta V

where

q is the magnitude of the charge of the particle

\Delta V is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J

- the potential difference is

\Delta V=4.50 V

So, the change in electrical potential energy is

\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J

7 0
4 years ago
A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ=cx2, where x is
grigory [225]

Answer:

Explanation:

λ=c x²

c = λ / x²

λ is mass / length

so its dimensional formula is ML⁻¹

x is length so its dimensional formula is L

c = λ / x²

= ML⁻¹ / L²

= ML⁻³

B )

We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M

The mass in the rod is symmetrically distributed on both side of middle point.

we consider a small strip of rod of length dx at x distance away from middle point

its mass dm = λdx = cx² dx

By integrating it from -L to +L we can calculate mass of whole rod , that is

M = ∫cx² dx

= [c x³ / 3] from -L/2 to +L/2

= c/3 [ L³/8 + L³/8]

M = c L³/12

c = 12 M L⁻³

C ) Moment of inertia of rod

∫dmx²

= ∫λdxx²

= ∫cx²dxx²

= ∫cx⁴dx

= c x⁵ / 5 from - L/2 to L/2

= c / 5 ( L⁵/ 32 +L⁵/ 32)

= (2c / 160)L⁵

= (c / 80) L⁵

= (12 M L⁻³/80)L⁵

= 3/20 ML²

=

=

4 0
3 years ago
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