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Furkat [3]
3 years ago
7

6. What is the change in temperature of a metal rod that is 55.0 cm long, decreases length by 0.20 cm, and that has a coefficien

t of linear expansion of 12 x 10-6 /C?
Physics
1 answer:
MrMuchimi3 years ago
4 0

Explanation:

We have,

Length of a metal rod is 55 cm or 0.55 m

Change in length is 0.2 cm or 0.002 m

It is required to find the change in temperature of a metal rod. The coefficient of linear expansion is given by :

\alpha =\dfrac{\Delta L}{L_0\Delta T}

\Delta T is the change in temperature

\Delta T =\dfrac{\Delta L}{L_0\alpha }\\\\\Delta T =\dfrac{0.002}{0.55\times 12\times 10^{-6}}\\\\\Delta T= 303.03^{\circ} C

So, the change in temperature is 303.03 degrees Celsius.

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A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

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p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

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3 years ago
Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?
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If an object is thrown upward with an initial velocity of 128 ​ft/second, then its height after t seconds is given by the follow
IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

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Given that,

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Firstly we need to calculate the time

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From equation (I)

\dfrac{dh}{dt}=128-64t

128-64t=0

t=\dfrac{128}{64}

t=2\ sec

Now, for maximum height

Put the value of t in equation (I)

h =128\times2-32\times4

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(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

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Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0
3 years ago
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