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: This means that the change in internal energy of the gas is equal to the heat exchanged: . However, since the gas is cooled, this means that Q is negative (heat is removed from the gas), so is negative, so the temperature of the gas, T, is decreased. Now if we look at the ideal gas equation
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:
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.
IS IT A MULTI CHOICE?? THEN SAY SO!!!
2 and 5 are the 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
We know the bat applies a force on the ball of F=400 N during t=0.02 sec, thus the impulse is
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.
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
B:science
C: scientific methods
D: technology
E: controlled variables
Controlled variables are the factor that remains at the same phases of an experiment.
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.
Variables can be controlled either directly by maintaining their value throughout a study (for example, by maintaining a constant room temperature in an experiment) or indirectly by using techniques like randomization or statistical control (e.g., to account for participant characteristics like age in statistical tests).
Hence, controlled variables are that variables which are kept at the same phases of an experiment.
To learn more about controlled experiment, refer the link:
#SPJ6
the controlled varaibles aka E. because the control group is Group of subjects in the experiment that experimental group as compared to
All of the other choices.
It's a graph of a vehicle's fuel efficiency ... what operating speeds make it use more or less gas to go 1 mile.