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3 years ago

Derive the dimension of coefficient of linear expansivity

2 answers:

5 0

Q1. A metal rod is of length 64.576 cm at a temperature 90°C whereas the same metal rod has a length of 64.522 cm at a temperature 12°C. Calculate the coefficient of linear expansion.

Reil [10]3 years ago

3 0

Answer:

The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.

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I) Give two practical<br>application of static friction<br>

bulgar [2K]

1). Walking / Driving

If there were no static friction between the soles of your shoes and the ground, then you could move your feet back and forth but your body would never go anywhere.

Same for using tires to move a car, a bus, a bicycle or a motorcycle.

2). Sleeping

If there were no static friction between your jammies and the sheet, you would slide right off of the bed whenever there was the slightest breeze of air in the room.

7 0

3 years ago

a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the we

andreev551 [17]

Explanation:

The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.

Sum the torques about the wedge (it may help to draw a diagram first). Take counterclockwise to be positive.

∑τ = Iα

W (0.25 m) − (100 N) (0.50 m) = 0

W = 200 N

Sum the forces in the y direction.

∑F = ma

F − 100 N − 200 N = 0

F = 300 N

8 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,

valina [46]

Answer:

$=2.99\times10^9J$

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement $S = 0.72\times10^3m$

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19° ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

$= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)$

$=2.99\times10^9J$

6 0

3 years ago

Two different simple harmonic oscillators have the same natural frequency (f=8.80 Hz) when they are on the surface of the Earth.

bekas [8.4K]

Answer:

8.80 Hz

Explanation:

The frequency of a loaded spring is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.

Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.

On the other hand, the frequency of the simple pendulum is affected because it is given by

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

$f_2 = 8.80\sqrt{\dfrac{1.67}{9.81}}=3.63 \text{ Hz}$

3 0

3 years ago

The formula can be used to convert temperatures between degrees Fahrenheit () and degrees Celsius (). How many degrees are in th

Jlenok [28]

Answer:

-30°C

Explanation:

F-32/180 =C-0/100

or, -22-32/180=C/100

or, -54/180*100=C

or, -0.3*100=C

therefore, C= -30

-22°F = -30°C

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8 0

3 years ago