Answer:
Temperature : 92.9 F
Internal Energy change: -2.53 Btu/lbm
Explanation:
As
mh1=mh2
h1=h2
In table A-11 through 13E
p2=120Psi, h1= 41.79 Btu/lbm,
u1=41.49
So T1=90.49 F
P2=20Psi
h2=h1= 41.79 Btu/lbm
T2= -2.43F
u2= 38.96 Btu/lbm
T2-T1 = 92.9 F
u2-u1 = -2.53 Btu/lbm
Answer:
m = 3 kg
The mass m is 3 kg
Explanation:
From the equations of motion;
s = 0.5(u+v)t
Making t thr subject of formula;
t = 2s/(u+v)
t = time taken
s = distance travelled during deceleration = 62.5 m
u = initial speed = 25 m/s
v = final velocity = 0
Substituting the given values;
t = (2×62.5)/(25+0)
t = 5
Since, t = 5 the acceleration during this period is;
acceleration a = ∆v/t = (v-u)/t
a = (25)/5
a = 5 m/s^2
Force F = mass × acceleration
F = ma
Making m the subject of formula;
m = F/a
net force F = 15.0N
Substituting the values
m = 15/5
m = 3 kg
The mass m is 3 kg
Answer:
0.25m²
Explanation:
We know that the summation of forces in the vertical direction is zero
So
PA-mg=0
A=mg/p
So
Substituting
A= 75* 9.8/3*10^-3
=0.25m² which is the total shoe area
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Answer:
1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>
2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>
3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>
4. The magnitude of the force on the object is a maximum on <em>the Top.</em>
Explanation:
<em>1. Because the change in position delta X is zero.</em>
<em>2. Because of delta X.</em>
<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>
<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>
If you soak this egg shell in vinegar (which is about 4% acetic acid), you start a chemical reaction that dissolves the calcium carbonate shell. The acetic acid reacts with the calcium carbonate in the egg shell and releases carbon dioxide gas that you see as bubbles on the shell.