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Tema [17]
2 years ago
12

An acetal polymer block is fixed to the rigid plates at its top and bottom surfaces. If the top plate displaces 2 mm horizontall

y when it is subjected to a horizontal force P = 2 kN ,determine the shear modulus of the polymer. The width of the block is 100 mm. Assume that the polymer is linearly elastic and use small angle analysi
Physics
1 answer:
Taya2010 [7]2 years ago
3 0

Answer:

The shear modulus of the polymer is 5 MPa

Solution:

As per the question:

displacement of the top plate, d = 2 mm = 0.002 m

Horizontal Force, P = 2 kN = 2000 N

Width of the block, w = 100 mm = 0.1 m

By small angle analysis:

Angle, \gamma = \frac{2}{200} = 0.01 rad

Now,

Stress, \tau = \frac{P}{Area} = \frac{2000}{0.4\times 0.1} = 50000 Pa

G = \frac{\tau}{\gamma} = \frac{50000}{0.01} = 5 MPa

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Answer: 71.93 *10^3 N/C

Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:

∫E*dr=Q inside/εo  Q inside is given by: λ*L then,

E*2*π*r*L=λ*L/εo

E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C

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How to convert 40m/min to km/h? need step by step explanation​
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Answer:

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A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N push
atroni [7]

Answer:

0.8214 m/s^2

Explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N      Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N    mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2

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3 years ago
A 10 ohms resistor is powered by a 5-V battery. The current flowing<br> through the source is:
mario62 [17]
  • Resistance=R=10ohm
  • Voltage=V=5V
  • Current=I

Applying ohm's law

\\ \sf\longmapsto \dfrac{V}{I}=R

\\ \sf\longmapsto I=\dfrac{V}{R}

\\ \sf\longmapsto I=\dfrac{5}{10}

\\ \sf\longmapsto I=0.5A

4 0
2 years ago
The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

8 0
3 years ago
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