The following
are the answers to the questions presented:
a. The joules of energy required to run a 100W light bulb for one day is 8640000J
b. The amount of coals that has to be burned to light that light bulb for one day is 0.96kg
The solution would
be like this for this specific problem:
<span>P=<span>W/s</span>→W=Pt=100W1day <span><span>24h/</span><span>1day </span></span><span><span>3600s/</span><span>1h</span></span>=8640000J</span>
<span>W=<span>30/100</span>wm→m=<span><span>100W/</span><span>30w</span></span>=<span><span>100×8640000J/</span><span>30×30×<span>10in thepowerof6 </span><span>J/<span>kg</span></span></span></span>=0.96kg</span>
<span>I am hoping that
these answers have satisfied your queries and it will be able to help you in
your endeavors, and if you would like, feel free to ask another question.</span>
Answer: B
Explanation: The rocket is accelerating the the spacecraft. Accelerating relates to Newtons Second Law of Motion.
(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
<u>The given parameters include:</u>
- constant velocity of the elevator, u₁ = 10 m/s
- initial velocity of the ball, u₂ = 20 m/s
- height of the boy above the elevator floor, h₁ = 2 m
- height of the elevator above the ground, h₂ = 28 m
To calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.

To learn more about projectile calculations please visit: brainly.com/question/14083704
A geological fold<span> occurs when one or a stack of originally flat and planar surfaces, such as sedimentary strata, are bent or curved as a result of permanent deformation.
So A fold is a Bend? in a rock. Maybe.
</span>A fault<span> is a planar fracture or discontinuity in a volume of </span>rock<span>, across which there has been significant displacement as a result of </span>rock<span>-mass movement.</span>