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2 years ago

6

Suppose that 2.5 moles of an ideal gas are in a chamber in equilibrium at temperature 310 K and volume 0.5 m3. 1) What is the pr

essure in the chamberi?

1 answer:

Basile [38]2 years ago

3 0

Answer:

The pressure in the chamber are of p= 0.127 atm.

Explanation:

n= 2.5 moles

T= 310 K

V= 0.5 m³ = 500 L

R= 0.08205746 atm. L /mol . K

p= n*R*T/V

p= 0.127 atm

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An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an e

Oksi-84 [34.3K]

Answer:

0.69 ohm

Explanation:

Heat generated per second, H = 50 cal/s

Potential difference, V = 12 V

Let R is the resistance of coil.

The formula for the heat is given by

$H = \frac{V^{2}}{R}t$

$50\times 4.186 = \frac{12^{2}}{R}\times 1$

R = 0.69 ohm

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2 years ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

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2 years ago

Two wheels are identical but wheel b is spinning with twice the angular speed of wheel

Murljashka [212]

<span>If two wheels are exactly the same but spin at different speeds, wheel b is twice te speed of wheel a, it is possible to find the ratio of the magnitude of radial acceleration at a singular point of the rim on wheel be to the spot is four.</span>

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2 years ago

A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed

Murrr4er [49]

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

$K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J$

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

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1 year ago

Q13. An electric bulb draws a current of 0.8A and works on 250V on the average 8 hours a day, if energy costs Rs 3 per kWh. Calc

olga_2 [115]

Using p=v * i
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2 years ago