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3 years ago

When a system does work, what happens to its internal energy

2 answers:

8 0

The first law, which deals with changes in the internal energy, thus becomes 0 = Q - W, so Q = W.

If the system does work, the energy comes from heat flowing into the system from the reservoir; if work is done on the system, heat flows out of the system to the reservoir

Ipatiy [6.2K]3 years ago

8 0

Answer:

Internal energy decreases

Explanation:

According to first law of thermodynamics heat can neither be created nor destroyed , but it changes from one form to another form.

Mathematical representation :

U=Q-w

Where U=Internal energy of system

w=Work done

Q=Heat gain or removed

When system does work then it take positive .When system does work then Q is positive .

Hence, the internal energy will decrease .

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Match each term with its best description.

Oxana [17]

The answers are:
1. Compaction - grains move closer together.
2. Foliation - parallel alignment of platy grains.
3. Recrystallization - grains take on a larger size.
4. Burial metamorphism - when a rock changes because of surrounding pressure.

Compaction produce rocks that have a tighter grain arrangement due to heat and pressure the rock undergoes. This process reduces the size of the rock.

Foliation occurs when pressure is put on one side of the grains more than any other direction. This creates platy structured rocks, which align under pressure.

Recrystallization is a process where the chemical composition a rock is changed. The mineral grains that are produced in this process are larger than the original grains.

Burial metamorphism occurs when rocks get buried deeper and deeper. They change due to the pressure from the overlying layers.

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4 years ago

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Select all of the answers that apply.

galben [10]

<span>The answers are --

a) wind direction
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f) location of precipitation</span>

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4 years ago

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Oxygen is a reactant in a combination reaction, which always produces energy in the form of heat and light. A. True B. False

harkovskaia [24]

The answer is B.False

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3 years ago

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Calculate the pressure exerted on the ground when a woman wears high heals. Her mass is 65 kg. and the area of each heal is 1 cm

erma4kov [3.2K]

Answer:

318.5 x 10^4 Pa

Explanation:

weight of woman = m g = 65 x 9.8 = 637 N

Area of both the heels = 1 x 2 = 2 cm^2 = 2 x 10^-4 m^2

Pressure is defined as the thrust acting per unit area.

P = F / A

Where, F is the weight of the woman and A be the area of heels

P = 637 / (2 x 10^-4) = 318.5 x 10^4 Pa

8 0

3 years ago

Un coche de carreras de Fórmula 1 (véase Figura 2.30) acelera desde el reposo a razón de 18 m.s-2 . Suma que se mueve en línea r

Lynna [10]

Answer:

I will answer in English.

Ok, we know that the acceleration is a = 18m/s^2, and we have that the initial velocity and position are both zero. (because it starts at rest)

then we have:

a(t) = 18m/s^2

for the velocity, we integrate over time (because the initial velocity is equal to zero we do not have any integration constant)

v(t) = (18m/s^2)*t

for the position we integrate again over time, and again, we do not have any integration constant

p(t) = (1/2)(18m/s^2)*t^2 = (9m/s^2)*t^2

a) The speed at t= 3s can be found by replacing t = 3s in the velocity equation.

v(3s) = (18m/s^2)*3s = 54m/s

b) the distance traveled by this time can be found by replacing t = 3s in the position equation.

p(3s) =(9m/s^2)*(3s)^2 = 81 m

c) first, we need to find what is the time when the position is equal to 200m.

p(t) = 200m = (9m/s^2)*t^2

√(200/9) s = t = 4.7s

Now we replace that time in the velocity equation and we get:

v (4.7s) = (18m/s^2)*4.7s = 84.6m/s

d) ok, to do this we know that.

1 hour has: 60*60 = 3600 seconds.

then we have the transformation k = 1h/3600s

1 km has 1000 meters.

then we have the transformation c = 1km/1000m

so we have that:

84.6m/s = 84.6m/s*(c/k) = 84.6*(3600/1000)km/h = 304.56 km/h

6 0

3 years ago