Define 1 pascal pressure
1 answer:
You might be interested in
We know that d=m/v where d is density, m is mass and v is volume. Lets transform formula to get v in times of d and m
Now we have to substitute our data to formula
V=
ml - its the answeer
Now we have to substitute our data to formula
V=
Answer:
The work energy added to the gas is
W = -72.48472 kJ
Explanation:
For constant pressure process
Work is given W = p×
However dH = m× = 1875 J
For mass m, dQ = 2625 J
That is m×1875 = 2625 or m = 2625/1875 = 1.4 kg
Therefore work doneper unit mass of gas is -W = p×(v₂ - v₁)
but pv = RT therefore work done per unit mass = -R×(T₂ - T₁) = -R×ΔT and
ΔT = 180.4K
Work done = -1.4×0.287×180.4 = -72.48472 kJ
a) 73.5 N
b) 551.3 J
c) -551.3 J
d) 0 J
e) 0 J
f) 0 J
g) 0 J
Explanation:
a)
There are two forces acting on the crate:
- The push of the worker, F, in the forward direction
- The frictional force, , in the backward direction, where
is the coefficient of friction
m = 50 kg is the mass of the crate
is the acceleration due to gravity
According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:
However, the crate here is moving with constant velocity, so its acceleration is zero:
So the previous equation becomes:
And we can find the magnitude of the applied force:
b)
The work done by the applied force on the crate is
where:
F is the magnitude of the force
d is the displacement of the crate
is the angle between the direction of the force and of the displacement
Here we have:
F = 73.5 N
d = 7.5 m
(the force is applied in the same direction as the displacement)
Therefore,
c)
The work done by friction on the crate is:
where in this case:
is the magnitude of the force of friction
d = 7.5 m is the displacement of the crate
, because the displacement is forward and the force of friction is backward, so they are in opposite direction
Therefore, the work done by the force of friction is:
d)
To find the normal force, we analyze the situation of the force along the vertical direction.
We have two forces on the vertical direction:
- The normal force, N, upward
- The force of gravity, , downward, where
m = 50 kg is the mass of the crate
is the acceleration due to gravity
Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:
The work done by the normal force is:
In this case, , since the normal force is perpendicular to the displacement of the crate; therefore, the work done is
e)
The work done by the gravitational force is:
where:
is the gravitational force
d = 7.5 m is the displacement of the crate
is the angle between the direction of the gravitational force (downward) and the displacement (forward)
Therefore, the work done by gravity is
f)
The total work done on the crate can be calculated by adding the work done by each force:
where we have:
is the work done by the applied force
is the work done by the frictional force
is the work done by the normal force
is the work done by the force of gravity
Substituting,
So, the total work is 0 J.
g)
According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:
where
W is the work done on the crate
is the change in kinetic energy of the crate
In this problem, we have:
(total work done on the crate is zero)
Therefore, the change in kinetic energy of the crate is: