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Musya8 [376]
3 years ago
6

An automobile tire having a temperature of −1.6 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 22 lb/in2 . What

would be the gauge pressure in the tire when its temperature rises to 45◦C? For simplicity, assume that the volume of the tire remains constant, that the air does not leak out and that the atmospheric pressure remains constant at 14.7 lb/in2 . Answer in units of lb/in2 .
Physics
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

25.8 lb/in²

Explanation:

Gay-Lussac's law tells us that given an ideal gas of a certain mass has a constant volume, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} } \\\\\frac{22}{-1.6+273.15} =\frac{P_{2} }{45+273.15} \\\\P_{2} = \frac{22*318.15}{271.55} = 25.8lb/in^{2}

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2.A chef leaves a copper spoon, a wooden spoon, and a steel spoon in a pot of boiling soup for several minutes. Which spoon shou
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where as heat transfer take place in steel as well as copper and chef will not be able to lift the spoon bare hand.

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Calculate the most charge that a 100pF capacitor (with a 1.0mm plate separation) can store if the capacitor breaks down with an
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Answer:

0.6 μC

Explanation:

C = capacitance of the capacitor = 100 x 10⁻¹² F

d = separation between the plates of capacitor = 1 mm = 1 x 10⁻³ m

E = Electric field = 6 x 10⁶ N/C

Q = Amount of charge

V = Potential difference

Potential difference is given as

V = E d

Amount of charge stored is given as

Q = CV

hence

Q = C E d

inserting the values

Q = (100 x 10⁻¹²) (6 x 10⁶) (1 x 10⁻³)

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FIGURE 2 shows a constant force, F of magnitude 20N pulled a light cord wrapped around a pulley to lift a bucket of mass 1.53kg.
slava [35]

(a) 14.3 rad/s^2

The angular acceleration of the pulley can be found by using the equivalent of Newton's second law for rotational motions:

\tau = I \alpha (1)

where

\tau is the net torque on the pulley

I=0.385 kg m^2 is the moment of inertia of the pulley

\alpha is the angular acceleration

First, we need to find the net torque. The torque exerted by the force F (forward) is:

\tau_t = F r = (20 N)(0.330 m)=6.6 Nm

While the frictional torque (backward) is \tau_f = 1.1 Nm. So, the net torque is

\tau = \tau_t - \tau_f=6.6 - 1.1 = 5.5 Nm

Now, re-arranging eq.(1), we find the angular acceleration:

\alpha = \frac{\tau}{I}=\frac{5.5}{0.385}=14.3 rad/s^2

(b) 4.72 m/s^2

Assuming that the string holding the bucket is inextensible, the bucket should have the same linear acceleration of a point on the edge of the pulley, which is given by

a= \alpha r

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Substituting into the equation, we find

a=(14.3)(0.330)=4.72 m/s^2

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