The strength of the friction doesn't matter. Neither does the distance or the time the asteroid takes to stop. All that matters is that the asteroid has
1/2 (mass) (speed squared)
of kinetic energy when it lands, and zero when it stops.
So
1/2 (mass) (original speed squared)
is the energy it loses to friction in order to come to rest.
<span>Answer:
Let m = mass of cannon
Then
10000 = ma
a = 10000/m
v^2 = u^2 + 2as
v^2 = 0 + 2as
84^2 = 2(2.21)(10000/m)
84^2 m = 4.42(10000)
m = 6.264172336
= 6.26 kg
Part 2
Range = u^2sin(2x38)/g
= 84^2sin(76)/9.8
= 698.6129229
= 698.6 m</span>
Answer:
A
Explanation:
Because D is definitely true and there is only one false sentence what means if that non of B or C is false because if one is false so other one needs to be too.
Answer: option d: The nucleus of Atom Q is more stable than the nucleus of Atom P.
Explanation:
Atom P is radioactive and disintegrates, it emits beta particles (high speed electrons or positrons) because it is not stable. On disintegration, it forms a stable Atom Q which is non-radioactive and thus it does not disintegrates further.
Thus, the correct option is only d. The nucleus of Atom Q is more stable than the nucleus of Atom P.
To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,








PART A)
Work done by the heat engine in each cycle = W



According to the value given we have that,


Power is defined as the variation of energy as a function of time therefore,




Therefore the interval for each cycle is 0.75s