Answer:
D) The acidic tomato juice reacted unfavorably with the aluminum pan
Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus,
=
= 171.82 kJ/kg
We know COP of heat pump
COP = 
= 
= 6.076
Therefore, Work out put, W = 
= 171.82 / 6.076
= 28.27 kJ/kg
Add the KE increase and the work done against friction.
The final velocity is twice the average, or 3.0 m/s
The final KE is (1/2)*25*3^2 = 112.5 J
The friction work done is 6*3.8 = 22.8 J
hope this is correct
Answer:
Approximately
, assuming that the acceleration of this ball is constant during the descent.
Explanation:
Assume that the acceleration of this ball,
, is constant during the entire descent.
Let
denote the displacement of this ball and let
denote the duration of the descent. The SUVAT equation
would apply.
Rearrange this equation to find an expression for the acceleration,
, of this ball:
.
Note that
and
in this question. Thus:
.
Let
denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is
, the net external force on this ball would be
.
Since
and
, the net external force on this ball would be:
.
·The acceleration of gravity is proportional to
1 / (the square of the distance from the center) .
When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is 9.8 m/s² .
The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)
= 490 newtons .
At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is
(9.8 m/s²) · (1/5)² = 0.39 m/s² .
The boy's weight is (mass) · (gravity) = (50kg) · (0.39 m/s²)
= 19.6 newtons .
Just as we expected, his weight at that distance is
(19.6 / 490) = 0.04 = 1/25 = 1/5² of his weight on the surface.