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Vsevolod [243]
3 years ago
15

A hole is drilled in a metal plate. When the metal is raised to a higher temperature, what happens to the diameter of the hole?

Physics
1 answer:
zysi [14]3 years ago
4 0

Answer:

The diameter of the hole increases

Explanation:

Metals expand and contract with temperature. Whenever metal is heated, it usually expands in relation to its thermal expansivity. This expansion leads to a slight increase in surface area.

Once the surface area of the metal changes, this means that the dimensions of the whole metal surface changed. As a result, the diameter of the hole drilled in the metal plate will change also. In our case, the diameter of the hole will increase.

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1.Mention two uses of the concave mirror.
Anuta_ua [19.1K]

Answer:

1. telescope

2.

f =  \frac{r}{2}

f =  \frac{r}{2}f- focal length

f =  \frac{r}{2}f- focal length r- the radius of curvature of the mirror

\frac{1}{f}  =   \frac{1}{p}  +  \frac{1}{l}

p-the distance of the object from the vertex of the mirror

l-the distance of the figure from the vertex of the mirror

8 0
3 years ago
Describe how refraction concentrates wave energy so that some parts of a beach erode more.
Zigmanuir [339]

Refraction is the bending of the waves which is result of the fact that different parts of the wave reach the water with different speeds because of the angle approaching the shore.

<span>The wave refraction disperses the wave energy in quiet water areas and sand is deposited.<span>  </span></span>



3 0
3 years ago
The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 f
Andrej [43]

Answer:

35.14°C

Explanation:

The equation for linear thermal expansion is \Delta L = \alpha L_0\Delta T, which means that a bar of length L_0 with a thermal expansion coefficient \alpha under a temperature variation \Delta T will experiment a length variation \Delta L.

We have then \Delta L = 0.481 foot, L_0 = 1671 feet and \alpha = 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a \Delta T =\frac{\Delta L }{\alpha L_0} = 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.

Since our temperature on a cool spring day was 13.0°C, our new temperature must be T_f=T_0+\Delta T = 35.14°C

3 0
3 years ago
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0
3 years ago
How much water can be held by a cylindrical tank with a radius of 12 feet and a height of 30 feet? A. 2,260.8 cubic feet B. 13,5
MakcuM [25]

Answer:

Volume, V = 13564.8  cubic feet

Explanation:

It is given that,

Radius of the cylindrical tank, r = 12 feet

Height of the tank, h = 30 feet

We need to find the water that can be held by a cylindrical tank i.e. we need to find the volume of the tank. It is given by :

V=\pi\times r^2\times h

V=3.14\times (12)^2\times 30

V = 13564.8  cubic feet

So, the water held by the tank is 13564.8  cubic feet. Hence, this is the required solution.

6 0
3 years ago
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