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3 years ago

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A bicycle tire has a pressure of 6.9 × 105 Pa at a temperature of 17.5°C and contains 2.00 L of gas. show answer Incorrect Answe

r What will its pressure be, in pascals, if you let out an amount of air that has a volume of 95 cm3 at atmospheric pressure and at the temperature of the tire? Assume tire temperature and volume remain constant. P2 = 14526315.79| sin() cos() tan() cotan() asin() acos() atan() acotan() sinh() cosh() tanh() cotanh() Degrees Radians π ( ) 7 8 9 HOME E ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 8% Potential 92% Submissions Attempts remaining: 1 (4% per attempt) detailed view 1 4% 2 4%

1 answer:

-BARSIC- [3]3 years ago

4 0

Answer:

the final pressure in pascal = $6.30*10^5 \ \ Pa$

Explanation:

Given that:

initial pressure = $6.9*10^5 \ Pa$

Initial Temperature = 17.5 °C = (17.5 + 273) K = 290.5 K

Volume = 2.00 L = 2.00 × 10⁻³ m³

Initial volume of the air occupied by gas V' = 95 cm³ = 95× 10⁻⁶ m

Using the ideal gas temperature;

PV = nKT

where

K = Boltzmann constant = 1.38 × 10⁻²³

From above expression;

$n = \frac{PV}{KT}$

$n = \frac{6.9*10^5*2.00*10^{-3}}{1.38*10^{-23}*290.5}$

$n = 3.44*10^{23}$

The number of moles that were removed from the tire is calculated as

$\Delta \ n = \frac{P_{atm}*V'}{KT}$

where

$P_{atm}$ = atmospheric pressure = $1.013*10^5$

$\Delta \ n = \frac{1.013*10^5*95*10^{-6}}{1.38*10^{-23}*290.5}$

$\Delta \ n = 2.4*10^{21}$

The remaining number of moles after the release of gas is

$n_2 = n- \Delta n$

$n_2 = 3.44*10^{23} -2.4*10^{21}$

$n_2 = 3.416*10^{23}$

Using the expression $P_2 = \frac{n_2 KT}{V}$ to determine the final pressure:

$P_2 = \frac{3.416*10^{23}*1.38*10^{-23}*290.5 }{2.00*10^{-3}}$

$P_2 = 6.30*10^5 \ \ Pa$

Hence, the final pressure in pascal = $6.30*10^5 \ \ Pa$

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1) 148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

$W = \Delta U = (mg) \Delta h$

where

mg is the weight of the object

$\Delta h$ is the change in height

For the box in this problem,

mg = 185 N

$\Delta h = 0.800 m$

Substituting into the equation, we find:

$W=(185)(0.800)=148 J$

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

$W=Fd$

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F is the magnitude of the force

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d = 35 m is the displacement of the car

Substituting,

$W=(825)(35)=28875 J$

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

$W=Fd$

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

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3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

$F = mg$

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m = 0.180 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

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$F=(0.180)(9.8)=1.76 N$

Now we find the work done by gravity using the same formula applied before:

$W=Fd$

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$W=(1.76)(2.5)=4.4 J$

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

$\Delta h = 1.2 m$

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

$W=mg \Delta h$

Solving for m, we find

$m=\frac{W}{g \Delta h}$

And substituting the numerical values, we find the mass of the box:

$m=\frac{7000}{(9.8)(1.2)}=595.2 kg$

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

$W=mg \Delta h$

Where $\Delta h$ is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of $\Delta h$ is the same for both of them, so the work they have done is exactly the same.

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3 years ago

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

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Answer:

a) v_average = 11 m / s, b) t = 0.0627 s

, c) F = 7.37 10⁵ N

, d) F / W = 35.8

Explanation:

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v² = v₀² - 2 a x

The fine speed zeroes them

a = v₀² / 2x

a = 22²/2 0.69

a = 350.72 m / s²

The average speed is

v_average = (v + v₀) / 2

v_average = (22 + 0) / 2

v_average = 11 m / s

b) The average time

v = v₀ - a t

t = v₀ / a

t = 22 / 350.72

t = 0.0627 s

c) The force can be found with Newton's second law

F = m a

F = 2100 350.72

F = 7.37 10⁵ N

.d) the ratio of this force to weight

F / W = 7.37 10⁵ / (2100 9.8)

F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

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A resistor, inductor, and a battery are arranged in a circuit. The circuit has an inductance of L = 1 H and a resistance of 1.4

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Answer:

$\tau \approx 7.14 \times 10^{-4}s \approx0.714ms$

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