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3 years ago

On reducing the volume of gas at constant temperature the pressure of gas increases why

1 answer:

8 0

Boyle's law is a gas law, stating that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If volume increases, then pressure decreases and vice versa, when temperature is held constant.

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Longitudinal waves transfer energy ___________ to the direction of the wave motion.

viva [34]

Longitudinal waves transfer energy parallel to the direction of the wave motion

4 0

3 years ago

Read 2 more answers

33 POINTS How can I get the temperature? SOUND SPEED 340 m/s = 331 m/s + (0,6xTemperature)

inessss [21]

(340-331)/0.6 = temp

9/0.6

90/6

30/2

15 degrees

6 0

3 years ago

What is the force required to accelerate a baseball with a mass of 0.145 kilograms and an acceleration of 35.00 meters/second sq

Paladinen [302]

Answer:5.075N

Explanation:

Mass=0.145kg

Acceleration=35m/s^2

Force=mass x acceleration

Force=0.145 x 35

Force=5.075N

3 0

3 years ago

A piece of metal has a mass of 9.00 kg. If it displaces water that fills a container 10.0 cm x 10.0 cm x 10.0 cm, what is the ma

Vadim26 [7]

Answer:

9000 kg/m³

Explanation:

Density is mass per volume.

D = M / V

D = (9.00 kg) / (0.100 m × 0.100 m × 0.100 m)

D = 9000 kg/m³

7 0

3 years ago

A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b

nekit [7.7K]

<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=\frac{1}{T}\intop_{0}^{T}p(t)dt$

But:

$v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)$

So:

$P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}$

$P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}$

In terms of RMS values:

$V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI$

7 0

3 years ago