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MrMuchimi
3 years ago
13

An object is inside a room that has a constant temperature of 292 K. Via radiation, the object emits three times as much power a

s it absorbs from the room. What is the temperature (in kelvins) of the object? Assume that the temperature of the object remains constant.
Physics
1 answer:
maw [93]3 years ago
8 0

Answer: Maybe he has the heater on

Explanation: If there is that much temperature then it has to be from something heated.

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The acceleration due to gravity on Earth is 9.81 m/s² [down]. A hockey puck was launched from the ice with a velocity of 50.0 m/
Gwar [14]

Answer:

0 m/s.

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When an object is shot up into the air the only force acting on it is gravity while neglecting wind. As the gravity is acting on the object it keeps decelerating till its velocity becomes 0 after which it can go no further. The maximum height of the object is at this point.

So, at its maximum height the velocity of the puck will be zero.

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How often do industrial oxygen cylinders need to be tested
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39.Explain the working mechanism of a mercury barometer.
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One end is sealed to prevent air from disturbing the scale and measures. Therefore, a Mercury Barometer works on the principle of balancing the atmospheric pressure with the volume of mercury present in the device
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What are the STEPS to using a triple beam balance
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4 years ago
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

6 0
4 years ago
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