Answer:
decreases by 1/9
Explanation:
the equation of kinetic energy is
here, we see that Kinetic Energy and v (velocity) are on opposide sides of the equation, this means that they are directly proportional (when one increases, the other one must increase (assuming mass (m) is constant)): KE∝
since V is decreasing by factor of 1/3, simply plug it in for v^2 to get (1/3)^2,
this equals (1/9). So KE will become 1/9 of original value
Answer:
Using the equation of continuity:
A
1
v
1
=
A
2
v
2
0.08
(
10
)
=
A
2
(
225
)
A
2
=
3.55
×
10
−
3
m
2
Q
2
=
A
2
v
2
Q
2
=
3.55
×
10
−
3
×
225
Q
2
=
0.798
m
3
/
s
Explanation:
Steady Flow Energy Equation:
The steady flow energy equation is a representation of the first law of thermodynamics. It is the conservation of energy law for an open system. A nozzle is an open system in the context of thermodynamics. It is used to produce a high velocity by reducing its pressure.
The steady flow energy equation can be given by the following formula:
h
1
+
1
2
v
2
1
+
g
z
1
+
q
=
h
2
+
1
2
v
2
2
+
g
z
2
+
w
where 'h' is enthalpy, 'v' is velocity, 'z' is height, 'q' is the heat and 'w' is work.
h
=
C
p
d
T
Answer and Explanation:
Given:
initial temp,
T
1
=
400
0
C
initial Pressure,
p
1
=
800
k
P
a
Initial Velocity,
v
1
=
10
m
/
s
Final temp,
T
2
=
300
0
C
Final Pressure,
p
2
=
200
k
P
a
Rate of heat loss, Q = 25 KW
Inlet Area,
A
1
=
800
c
m
2
As per the steady flow energy equation:
h
1
+
1
2
v
2
1
+
g
z
1
+
q
=
h
2
+
1
2
v
2
2
+
g
z
2
+
w
Since, there is external work, w= 0. Also, consider there is a negligible change in KE.
h
1
+
1
2
v
2
1
+
q
=
h
2
+
1
2
v
2
2
h
1
−
h
2
+
1
2
v
2
1
+
q
=
1
2
v
2
2
C
p
(
T
1
−
T
2
)
+
1
2
(
10
)
2
+
25000
=
1
2
v
2
2
2
(
400
−
300
)
+
50
+
25000
=
1
2
v
2
2
2
(
400
−
300
)
+
50
+
25000
=
1
2
v
2
2
25250
=
1
2
v
2
2
v
2
≈
225
which is the answer.
Using the equation of continuity:
A
1
v
1
=
A
2
v
2
0.08
(
10
)
=
A
2
(
225
)
A
2
=
3.55
×
10
−
3
m
2
Now, volume flow rate,
Q
2
=
A
2
v
2
Q
2
=
3.55
×
10
−
3
×
225
Q
2
=
0.798
m
3
/
s
Answer:
1 The weight of the foundation block should be enough to withstand vibrations
2 The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.
3.The base should be sufficiently rigid to have the requisite rigidity
4 a distance should be created all across machine foundation to separate it from the adjacent parts of the building
Explanation:
1. The weight of the foundation block should be enough to withstand vibrations and to avoid friction between device and the surrounding soil as well. This can be done by increasing the base block weight in supporting with engine power
2.The foundation must be so dimensional that the resulting pressure due to weight of the device as well as the weight of foundation moves through the center of gravity of a base impact area in order to avoid the risk of specific settlement.
3.The base should be sufficiently rigid to have the requisite rigidity as the slightest misdirection of foundation may cause significant bearing disorders.
4.To avoid propagation of vibration from a device to the adjacent parts of the building, a distance should be created all across machine foundation to separate it from the adjacent parts of the building.
The answer is the third statement- Tundra, wetland, rainforest, desert