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tresset_1 [31]
3 years ago
5

Gary rides his bike down the street in 30 s. His speed was a constant 5 m/s. What is the value of his acceleration during this t

ime?
Physics
1 answer:
natta225 [31]3 years ago
3 0

constant speed =5m/s

time he travel=30sec

acceleration = 5/30

=0.167m/s^2

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Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
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Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

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Answer: The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

I = \sqrt{\frac{P}{R} }

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }

Then, simplify the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} }

Rationalize the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}

Simplify the numerator by finding its factors:

I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}

The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

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Correct answer: Kinetic energy of rotation, kinetic energy of translation and potential energy

Explanation:

Before it began to rotate it possessed potential energy. After being released from the rest it starts to roll on the roof and get kinetic energy of rotation, kinetic energy of translation along with potential energy.

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