The acceleration due to gravity on the surface of Jupiter is about 2.5 times the acceleration due to gravity on Earth’s surface.
1 answer:
Answer: The correct answer is option C.
Explanation:
Weight = Mass × Acceleration
Let the mass of the space probe be m
Acceleration due to gravity on the earth = g
Weight of the space probe on earth = W
Acceleration due to gravity on the Jupiter = g' = 2.5g
Weight of the space probe on earth = W'
The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.
Hence, the correct answer is option C.
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(a) No, because the mechanical energy is not conserved
Explanation:
The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:
(1)
However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.
Therefore, eq. (1) can be rewritten as
which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane () and part is lost because of the air resistance (
).
(b) 77.8 m/s
First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:
Now we can calculate the acceleration of the plane, by using Newton's second law:
where m is the mass of the plane.
Finally, we can calculate the final speed of the plane by using the equation:
where
is the final velocity
is the initial velocity
is the acceleration
is the distance travelled
Solving for v, we find
Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached
