Answer:
Explanation:
work done=force*displacement
=350N*15m
=5250 joule
Answer
D. move a small magnet back and forth within a section of the coiled wire.
Explanation:
i put that for the test and i got it right
-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.
-- The mass of the satellite makes no difference.
Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is
(3.95 x 10⁵m) + (4.2 x 10⁶m)
= (3.95 x 10⁵m) + (42 x 10⁵m)
= 45.95 x 10⁵ m
The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.
The bird completes a revolution every 2.0 hours,
so its speed in orbit is
(91.9 π x 10⁵ m) / 2 hr
= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)
= 0.04 x 10⁵ m/sec
= 4 x 10³ m/sec
(4 kilometers per second)
Answer:
charge on each
Q1 = 2.06 ×
C
Q2 = 7.23 × C
when force were attractive
Q1 = 1.07 × 
C
Q2 = -1.39 × C
Explanation:
given data
total charge = 93.0 μC
apart distance r = 1.14 m
force exerted F = 10.3 N
to find out
What is the charge on each and What if the force were attractive
solution
we know that force is repulsive mean both sphere have same charge
so total charge on two non conducting sphere is
Q1 + Q2 = 93.0 μC = 93 ×
C
and
According to Coulomb's law force between two sphere is
Force F = 
.........1
Q1Q2 = 
here F is force and r is apart distance and k is 9 × 
N-m²/C² put all value we get
Q1Q2 = 
Q1Q2 = 1.49 × 
C²
and
we have Q2 = 93 × C - Q1
put here value
Q1² - 93 × Q1 + 1.49 × = 0
solve we get
Q1 = 2.06 × C
and
Q1Q2 = 1.49 ×
2.06 × Q2 = 1.49 ×
Q2 = 7.23 × C
and
if force is attractive we get here
Q1Q2 = - 1.49 × C²
then
Q1² - 93 × Q1 - 1.49 × = 0
we get here
Q1 = 1.07 × C
and
Q1Q2 = - 1.49 ×
2.06 × Q2 = - 1.49 ×
Q2 = -1.39 × C
The Correct answer to number 1 is A or D
The correct answer to number 2 is C because transmission meaning is to travel.
PS. I think number 1 is D
