Fg=mg
Fg/g=m
m=(1900N) / (9.81N/kg)
m=193.68kg
<span>
Work = (magnitude of the force) x (distance in the direction of the force)
A 100-N force lifts an object 3 m. ==> (100 x 3) = 300 joules
B. A 250-N force moves an object 1 m. ==> (250 x 1) = 250 joules
C. A 25-N force moves an object 8 m. ==> (25 x 8) = 200 joules
D. A 600-N force is applied to an object
and moves it 0 m. ==> (600 x 0) = no work
</span> Winner ! Least work.
For a point charge, how does the potential vary with distance from the point charge, r?
a constant
b. r.
c. 1/r.
d.
.
e.
.
Answer:
The correct option is C
Explanation:
Generally for a point charge the electric potential is mathematically represented as

Here we can deduce that the electric potential varies inversely with the distance i.e

So
Answer:
Explanation:
Power P = V² / R
a ) The resistance is changed to 2.90R
Power will become 1 / 2.9 times .
b )The voltage of the battery is now 2.90V, but the resistance is R
P = (2.9V)² / R
= 8.41 x V² / R
So power becomes 8.41 times
c )The resistance is 2.90R and voltage is 2.90V
Power P = (2.9V)² / 2.9 R
= 2.9 V²/R
So power becomes 2.9 times
d ) The resistance is 2.90R and the voltage is V/2.90
Power P = ( V/2.90)² x 1 / 2.90R
1 / ( 2.9 )³ x V² / R
= 1 / 24.389 x V² / R
So power becomes 1 / 24.389 times .