Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
The answer is D.) all of the above
Answer:
The neutron loses all of its kinetic energy to nucleus.
Explanation:
Given:
Mass of neutron is 'm' and mass of nucleus is 'm'.
The type of collision is elastic collision.
In elastic collision, there is no loss in kinetic energy of the system. So, total kinetic energy is conserved. Also, the total momentum of the system is conserved.
Here, the nucleus is still. So, its initial kinetic energy is 0. So, the total initial kinetic energy will be equal to kinetic energy of the neutron only.
Now, final kinetic energy of the system will be equal to the initial kinetic energy.
Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.
Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.
Therefore, the neutron loses all of its kinetic energy to nucleus.
Answer:
Heat required to raise the temperature of the aluminium is 4750 J
Explanation:
As we know that the heat energy required to raise the temperature of the aluminium is given as

here we know that
m = 50 g


so we have

