Answer:
A) 1.88 * 10^17 m
B) 1.22 * 10^34 J
C) 1.95 * 10^34 J
Explanation:
Parameters given:
Mass of planet = 7.00 * 10^25 kg
Radius of orbit = 6.00 * 10^11 m
Force exerted on planet = 6.51 * 10^22 N
Velocity of planet = 2.36 * 10^4 m/s
A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).
The circumference of the orbit is
C = 2 * pi * R
R = radius of orbit
C = 2 * 3.142 * 6.0 * 10¹¹
C = 3.77 * 10¹² m
Hence, distance traveled will be:
D = 0.5 * 3.77 * 10¹²
D = 1.88 * 10 ¹² m/s
B) Work done is given as:
W = F * D
W = 652 * 10²² * 1.88 * 10¹¹
W = 1.22 * 10³⁴ J
C) Change in Kinetic energy is given as:
K. E. = 0.5 * m * v²
K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²
K. E. = 1.95 * 10³⁴ J
The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
a = g = -9.8 
downward (acceleration due to gravity).
By using Suvat equation:
v = u + at
where: v is the velocity at time t
u = 40.0 m/s is the initial velocity
a = g = -9.8
is the acceleration
To find the time t at which the velocity is v = 20.7 m/s
Therefore,

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
The complete question is:
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?
To learn more about uniformly accelerated motion refer to:
brainly.com/question/14669575
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Interval training is simply alternating short bursts (about 30 seconds) of intense activity with longer intervals (about 1 to 2 minutes) of less intense activity. For instance, if your exercise is walking and you're in good shape, you might add short bursts of jogging into your regular brisk walks.
Answer:
The current through the inductor at the end of 2.60s is 9.7 mA.
Explanation:
Given;
emf of the inductor, V = 41.0 mV
inductance of the inductor, L = 13 H
initial current in the inductor, I₀ = 1.5 mA
change in time, Δt = 2.6 s
The emf of the inductor is given by;

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.