Answer:

Explanation:
A closed system is a system where exists energy interactions with surroundings, but not mass interactions. If we neglect any energy interactions from boundary work, heat, electricity, magnetism and nuclear phenomena and assume that process occurs at steady state and all effects from non-conservative forces can be neglected, then the equation of energy conservation is reduce to this form:
(1)
Where:
- Change in kinetic energy of the system, measured in joules.
- Change in gravitational potential energy of the system, measured in joules.
If we know that
and
, then we get the following equation:
(2)
Where
and
stands for initial and final states of each energy component.
Hence, the right answer is 
Answer:
Explanation:
Given
mass of bus along with travelers travelling in North direction is 
speed of bus towards North 
mass of bus travelling in South direction is 
speed of bus 
mass of each Passenger in south moving bus 
Momentum of North moving bus



Momentum with south moving bus


For total momentum to be towards south
should be greater than 0
thus for least value of n



Answer:
a)906.5 Nm^2/C
b) 0
c) 742.56132 N•m^2/C
Explanation:
a) The plane is parallel to the yz-plane.
We know that
flux ∅= EAcosθ
3.7×1000×0.350×0.700=906.5 N•m^2/C
(b) The plane is parallel to the xy-plane.
here theta = 90 degree
therefore,
0 N•m^2/C
(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
therefore, applying the flux formula we get
3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C
Answer:

Explanation:
Given

Required
Rewrite using scientific notation
The format of a number in scientific notation is

Where 
So the given parameter can be rewritten as

Express as a power of 10

Hence, the equivalent of the mass of the sun in scientific notation is:
