Gas pressure within a fixed container will increase when _____.1) gas is cooled
2) gas is heated
4) gas volume changes
5) mass of a gas is increased

Answers

Answer 1
Answer:

Answer:

2) gas is heated

Explanation:

In this problem, we have a gas in a fixed container (so, the volume of the gas does not change). Let's see what happens in the different situations listed in the question.

1) Gas is cooled. Since the volume if fixed, the work done by the gas is zero: W=p\Delta V = 0 This means that the change in internal energy of the gas is equal to the heat exchanged: \Delta U = Q. However, since the gas is cooled, this means that Q is negative (heat is removed from the gas), so \Delta U is negative, so the temperature of the gas, T, is decreased. Now if we look at the ideal gas equation

pV=nRT

We see that since V, n and R remain constant, a decrease in T corresponds to a decrease in p: so, the pressure of the gas decreases.

2) Gas is heated. If we apply exactly the same reasoning of point 1), we see that when the gas is heated, its temperature increases: therefore, according to the ideal gas equation, its pressure must increase as well.

3) Gas volume change. By looking again at the ideal gas equation:

pV=nRT

We see that in this case we don't have enough information to determine how does the gas pressure change. In fact, we don't know whether the volume V increases or decreases, so we cannot determine how p changes.

5) mass of a gas is increased. An increase in the mass of the gas corresponds to an increase in the number of moles, n. However, in the ideal gas equation, we don't know if the temperature of the gas, T, remains the same or not: therefore, we don't have enough information to tell how does p change.

Answer 2
Answer:

IS IT A MULTI CHOICE?? THEN SAY SO!!!

2 and 5 are the answers


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How much force is needed to get a 0.9kg block of wood to slide if the static coefficient of friction is 0.8

Answers

0.8?? DONT know how to solve

a baseball bat is in contact with a ball for 0.02 seconds, and exerts a force of 400 newtons. what is the impulse on the ball?​

Answers

Answer:

I=8 Nw.s

Explanation:

Impulse

It is a physical magnitude that measures the force applied to an object by the time that force is acting. It produces a change in the momentum of the object, which is expressed in the following formula

I=\Delta p=m(v_2-v_1)=F.t

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. A projectile is fired with an initial velocity of 113 m słatan angle of 60.0degrees above the horizontal from the top of a cliff 49.0 m high.
Calculate:
(a) the time to reach maximum height
(b) the maximum height above the base of the cliff reached by the
projectile
(c) thetotal time it is in the air
(d) the horizontal range of the projectile.​

Answers

Answer:

a) 9.99 s

b) 538 m

c) 20.5 s

d) 1160 m

Explanation:

Given:

x₀ = 0 m

y₀ = 49.0 m

v₀ = 113 m/s

θ = 60.0°

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

a) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find t.

vᵧ = aᵧ t + v₀ᵧ

(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)

t ≈ 9.99 s

b) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find y.

vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)

(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)

y ≈ 538 m

c) When the projectile lands, y = 0 m.  Find t.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²

You'll need to solve using quadratic formula:

t ≈ -0.489, 20.5

Since negative time doesn't apply here, t ≈ 20.5 s.

d) When the projectile lands, y = 0 m.  Find x. (Use answer from part c).

x = x₀ + v₀ₓ t + ½ aₓ t²

x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²

x ≈ 1160 m

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Answers

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Answers

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Anything kept constant or constrained in a research study is referred to as a control variable. Despite not being relevant to the study's objectives, this variable is controlled because it might have an impact on the results.

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brainly.com/question/1537888

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