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salantis [7]
2 years ago
8

A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introdu

ction of 430 J of heat caused the system to expand, doing 238 J of work against a constant pressure of 101 kPa. What is the value of for this process
Physics
1 answer:
xeze [42]2 years ago
8 0

Answer: You do not specify what is being asked for. ∆E? ∆H?

∆E = (430 - 238) J = 192 J

∆H = 430 J

Explanation:

If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.

Therefore ∆H = 430 J

If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w

The question states that 238 J of work are done AND the system expanded

(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)

Therefore, ∆E = (430 - 238) J = 192 J

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What is a concentration?​
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Answer/Explanation:

In chemistry, concentration is the abundance of a constituent divided by the total volume of a mixture. Several types of mathematical description can be distinguished: mass concentration, molar concentration, number concentration, and volume concentration.

A concentration refers to a subject or study within a specific major. As part of the degree coursework, concentration classes count toward your major requirements as well.

8 0
2 years ago
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A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

6 0
3 years ago
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

3 0
3 years ago
I did questions 2 and 3 but I don't know if they are right. Someone help!
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You got it right my friend

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Psuedopod in a sentence
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