M= 100000 kg
a= 2m/s²
Use Newton's second law:
F=ma=100000kg • 2 m/s²
= 200000 N
Answer:
Explanation:
It is given that the sphere is insulated from ground and a large charge is placed on the sphere. The charge on the hollow sphere will always remain on the outer surface of the sphere and there will be no charge on the inner surface of the sphere.
If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.
If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.
If a person does not touches the sphere then the charge on the outer surface will be zero and there will be a positive charge on the inner surface of the sphere
Answer:
Our appliance and electronic energy use calculator allows you to estimate your ... using the formulas provided below; Installing a whole house energy monitoring system. ... energy consumption a few watt-hours, and you can use a monitor to estimate those too. ... Estimate the number of hours per day an appliance runs.
Explanation:
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
