Answer:
mass of the planet: 
Explanation:
When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):
where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:
Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.
We know the orbital radius R (
, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.
We know that the moon makes a full circumference (
) in 388800 seconds, therefore its tangential velocity is:
where we rounded the velocity to one decimal.
Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.
Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:
Answer:
a 15.22 m/s
b 45.65 m
Explanation:
Using the same formula,
x = vt, where
x is now 45.65, and
t is 3 s, then
45.65 = 3v
v = 45.65/3
v = 15.22 m/s
See the attachment for the part b. We used the distance gotten in part B, to find question A
Answer:
<em>b. The current in the loop always flows in a counterclockwise direction.</em>
<em></em>
Explanation:
When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where 
is the initial velocity.
(a).
When the projectile hits the 50m mark, 
; therefore,
solving for 
we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed 
of
Kinetic energy is calculated through the equation,
KE = 0.5mv²
At initial conditions,
m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J
m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J
Due to the momentum balance,
m₁v₁ + m₂v₂ = (m₁ + m₂)(V)
Substituting the known values,
(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)
V = 0.2977 m/s
The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J
The difference between the kinetic energies is 0.0473 J.
