Answer:

Explanation:
From the question we are told that
Nucleus diameter 
a 12C nucleus
Required kinetic energy 
Generally initial speed of proton must be determined,applying the law of conservation of energy we have

where
=initial kinetic energy
=final kinetic energy
=initial electric potential
=final electric potential
mathematically

where
=distance b/w charges
=nucleus charge 
=constant
=proton charge
Generally kinetic energy is know as

Therefore
Generally equation for radius is 
Mathematically solving for radius of nucleus


Generally we can easily solving mathematically substitute into v_1









Therefore the proton must be fired out with a speed of 
V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
The answer would be 27,000 Joules because (1/2) m v^2 =30*900 which equals 27,000 J
Answer:
Tension T1 is less than tension T2.
T1 < T2
Explanation:
According to given data,
mass of box A ( mA) is grater than mass of box B (mB)
we can write,
m(A) > m(B)
Newton's second law states that:
Tension of object is directly proportional to the mass of the system.
T ∝ m
here Boxes A and B are being pulled to the right on a frictionless surface,
so Tension T1 generates due to the mass of box A m(A)
and Tension T2 arises due to mass of the system m(A) + m(B)
Thus tension T1 will be less than tension T2
T1 < T2
learn more about Tension force here:
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Answer:
Zero
Explanation:
The work done by a force on an object is given by:

where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and the displacement of the object
In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so
and
. Therefore, the work done by gravity is also zero.