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Valentin [98]
3 years ago
5

The contents of the cylinder is then entirely transfered to a partially filled holding tank of volume 10,000 L originally at pre

sure 2 atm and 300 K. What is the new pressure in the holding tank at 300 K?
Physics
1 answer:
Dahasolnce [82]3 years ago
7 0

Answer:

The new pressure in the holding tank is 2 atm.

Explanation:

Given that,

Volume = 10000 L

Initial pressure 2 atm

Initial temperature = 300 k

We need to calculate the new pressure

Using equation of ideal gas

PV=RT

R=\dfrac{PV}{T}

Where,P = pressure

T = temperature

V = volume

R = constant

Put the value into the formula

Volume of gas are remain constant

So,

\dfrac{P_{i}V}{T_{i}}=\dfrac{P_{f}V}{T_{f}}

\dfrac{2\times10000}{300}=\dfrac{P_{f}\times10000}{300}

P_{f}=2\ atm

Hence, The new pressure in the holding tank is 2 atm.

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A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has
irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration (a) consist in two components, <em>radial</em> (a_{r}) and <em>tangential</em> (a_{t}), in meters per square second:

a_{r} = \omega^{2}\cdot r (1)

a_{t} = \alpha \cdot r (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

a = \sqrt{a_{r}^{2}+a_{t}^{2}} (3)

Where:

r - Radius of the wheel, in meters.

\omega - Angular speed, in radians per second.

\alpha - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

\omega = \omega_{o}+ \alpha\cdot t (4)

Where:

\omega_{o} - Initial angular speed, in radians per second.

t - Time, in seconds.

If we know that r = 0.1\,m, \alpha = 2\,\frac{rad}{s^{2}}, \omega_{o} = 0\,\frac{rad}{s} and t = 0.60\,s, then the total linear acceleration is:

\omega = \omega_{o}+ \alpha\cdot t

\omega = 1.2\,\frac{rad}{s}

a_{r} = \omega^{2}\cdot r

a_{r} = 0.144\,\frac{m}{s^{2}}

a_{t} = \alpha \cdot r

a_{t} = 0.2\,\frac{m}{s^{2}}

a = \sqrt{a_{r}^{2}+a_{t}^{2}}

a \approx 0.246\,\frac{m}{s^{2}}

The total linear acceleration is approximately 0.246 meters per square second.

3 0
3 years ago
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

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3 years ago
You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler
lbvjy [14]
In a stationary situation, the weight of person is
W=mg=(62 kg)(9.81 m/s^2) = 608.2 N
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
\sum F = ma
W-W'=ma
where a is the acceleration of the elevator. If we solve for a, we find
a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.
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