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Ronch [10]
3 years ago
14

How do I find angular speed of a rotating mass

Physics
1 answer:
kobusy [5.1K]3 years ago
8 0
Angular speed is omega in radians/sec.
tangential speed = radius time omega
so take the tan speed and divide by the radius
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What happens to light as it moves at an angle into a medium that has a lower index of refraction?
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It speeds up, and the angle increases.

Explanation:

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Answer each question yes or no. Assume the motions and currents mentioned are along the x axis and fields are in the y direction
Sergeeva-Olga [200]

(a)  Does an electric field exert a force on a stationary charged object? YES ( F = Eq)

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(d) Does a magnetic field do so? YES ( F = qvB)  

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2 years ago
16. Two capacitors have an equivalent
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Answer:

C1 + C2 = 30     parallel connection

C1 * C2 / (C1 + C2) = 7.2  series connection

C1 * C2 = 7.2 * (C1 + C2) = 216

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C2^2 + 216 = 30 C2

C2^2 - 30 C2 + 216 = 0

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3 years ago
What color bends the most
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A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa
aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:        

The expression for the power radiated is as follows;

P=A\epsilon\sigma T^{4}

Here, A is the area, \sigma is the stefan's constant,\epsilon is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}

Here, A_{1} is the area of the sphere, A_{2} is the area of the cube,T_{1}  is the temperature of the sphere and T_{2}  is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

A_{1}T_{1}^{4}=A_{2}T_{2}^{4}

The area of the sphere is A_{1}=4\pi \times r^{2}.

Here, r is the radius of the sphere.

The area of the cube is A_{2}=6\times a^{2}.

Here, a is the edge of the cube.

Put A_{1}=4\pi \times r^{2} and A_{2}=6\times a^{2}.

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}  ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

V_{1}=V_{2}

Here,V_{1},V_{1} are the volumes of the sphere and the cube.

\frac{4}{3}\pi r^{3}=a^{3}

\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}  

Put this value in the equation (1).

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}

Put T_{1}=500°c.

T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}

T_{2}=472.2^{\circ}c

Therefore, the value of the celsius temperature of the cube is 472.7°c.    

5 0
3 years ago
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