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luda_lava [24]
3 years ago
6

Explain how you can use Boyle's law to determine the new volume of gas when its pressure is increased from 270kPa to 540kPa? The

orginal volume of gas was 1L. Assume the temperature and number of particles are constant. What is the new volume?
Physics
1 answer:
ivann1987 [24]3 years ago
4 0
I hope this is what you are looking for P1 = 270 kPa V1 = 1 L P2 = 540 kPa V2 = ? Use Boyle's Law: P1V1 = P2V2 V2 = P1 V1 / P2 V2 = 270 x 1 / 540 V2 = 0.5 L
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A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is
AveGali [126]

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

Circular \; speed (V) = \frac {2 \pi r}{t}

Where;

  • r represents the radius and t is the time.

Substituting into the formula, we have;

Circular \; speed (V) = \frac {2*3.142*2600}{360}

Circular \; speed (V) = \frac {16338.4}{360}

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

Centripetal \; acceleration = \frac {V^{2}}{r}

Where;

  • V is the circular speed (velocity) of an object.
  • r is the radius of circular path.

Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

Centripetal \; acceleration = \frac {2059.34}{2600}

<em>Centripetal acceleration = 0.79 m/s²</em>

3 0
3 years ago
A Explain why the sound produced by the horn of an approaching car
aleksklad [387]
When a car approaches you, the sound waves that reach you have a shorter wavelength and a higher frequency. You hear a sound with a higher pitch. When the car moves away from you, the sound waves that reach you have a longer wavelength and lower frequency.

?? ⬇️

An approaching source moves closer during period of the sound wave so the effective wavelength is shortened, giving a higher pitch since the velocity of the wave is unchanged. Similarly the pitch of a receding sound source will be lowered.

The Doppler effect is an effect observed in light and sound waves as they move toward or away from an observer. One simple example of the Doppler effect is the sound of an automobile horn. Picture a person standing on a street corner. A car approaches, blowing its horn.

Comparing two waves of the same wavelength, a higher frequency is associated with faster movement. Comparing two waves of different wavelengths, a higher frequency doesn't always indicate faster movement, although it can. Waves of different wavelengths can have the same frequency.

The pitch of a sound is our ear's response to the frequency of sound. Whereas loudness depends on the energy of the wave. ... The pitch of a sound depends on the frequency while loudness of a sound depends on the amplitude of sound waves.
5 0
3 years ago
Question 5
Ksenya-84 [330]

Answer:

The force of the nail pushing in the opposite direction

5 0
3 years ago
A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra
blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

y=\sqrt{1+x^3}

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

Now at (2,3)

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s

7 0
3 years ago
Find the velociity of a car which travels 35 m to the right over a period of 40 seconds
GalinKa [24]

Answer:

the velocity of the car is 0.875 m/s

Explanation:

v = d \div t \\ v = 35m \div 40sec = 0.875 \: m \: per \: sec

therefore the V of car is 0.875 m

8 0
3 years ago
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