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Scilla [17]
3 years ago
15

A bicycle tire is inflated to a gauge pressure of 3.71 atm when the temperature is 16°C. While a man rides the bicycle, the temp

erature of the tire rises to 45°C. Assuming the volume of the tire does not change, find the gauge pressure in the tire at the higher temperature.
Physics
1 answer:
Contact [7]3 years ago
8 0

Answer:

The pressure of the tire will be 4.19 atm

Explanation:

The ideal gas equation is:

p0 * V0 / T0 = p1 * V1 / T1

Since V0 = V1 because the volume of the tire doesn't change

p0 / T0 = p1 / T1

p1 = p0 * T1 / T0

We need absolute temperatures for this equation

16 C = 289 K

45 C = 318 K

Now we replace the values:

p1 = 3.71 * 318 / 289 = 4.19 atm

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Alexxx [7]

Answer:

You didn't give the information needed for the answer bud

Explanation:

8 0
3 years ago
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
Wittaler [7]

Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

6 0
3 years ago
You are standing on the ground. Describe two ways that you could increase the pressure between your feet and the ground.
ch4aika [34]

Answer:

1:by standing with one feet

2:by standing with your toes

Explanation:

1: by standing with one feet

2:by standing with ur toes

Pressure is inversely proportional to area, the little the area, the more the pressure.the bigger the area,the lesser the pressure

3 0
4 years ago
What are the possible units for half life?
lilavasa [31]

Answer:

Half-life units in 'years' & 'seconds' are the most widely used, but actually, whatever you wish ... sec, min, hrs, days, yrs. The application and one's preference could dictate more appropriate units.

Explanation:

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Tpy6a [65]
Electric force is force between 2 charges seperated by distance in space.While  gravitational force has to be dependent on the independent action.<span />
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3 years ago
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