Answer:
500J
Explanation:
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A solar eclipse will be visible over a wide area of the north polar region
on Friday, March 20.
England is not in the path of totality, but it's close enough so that a large
part of the sun will be covered, and it will be a spectacular sight.
For Londoners, the eclipse begins Friday morning at 8:25 AM,when the
moon just begins to eat away at the sun's edge. It advances slowly, as more
and more of the sun disappears, and reaches maximum at 9:31 AM. Then
the obscured part of the sun begins to shrink, and the complete disk is
restored by the end of the eclipse at 10:41AM, after a period of 2 hours
16 minutes during which part of the sun appears to be missing.
The catch in observing the eclipse is:
<em><u>YOU MUST NOT LOOK AT THE SUN</u></em>.
Staring at the sun for a period of time can cause permanent damage to
your vision, even though <em><u>you don't feel it while it's happening</u></em>.
This is not a useful place to try and give you complete instructions or
suggestions for observing the sun over a period of hours. Please look
in your local newspaper, or search online for phrases like "safe eclipse
viewing".
Answer:
4.2 is the answer
Explanation
The image formed in a plane mirror is an equal distance behind the mirror as the object in front of it.
Step 1: the equation to this problem would be: 8.4/2
Step 2: 8.4 ÷ 2 = 4.2
The average power is 
Explanation:
First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:
where
:
is the final kinetic energy of the car, with
m = 2000 kg is the mass of the car
v = 60 m/s is the final speed of the car
is the initial kinetic energy of the car, with
u = 30 m/s is initial speed of the car
Soolving:
Now we can find the power required for the acceleration, which is given by

where
t = 9 s is the time elapsed
Solving:

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Answer:
Explanation:
Area A of the coil = .1 x .1 = .01 m²
no of turns n = 5
magnetic field B = .5 t²
Flux Φ perpendicular to plane passing through it.= nBA sin30
rate of change of flux
dΦ/dt = nAdBsin30 / dt
= nA d/dt (.5t²x .5 )
= nA x 2 x .25 x t
At t = 4s
dΦ/dt = nA x 2
= 5x .01 x 2
= .1
current = induced emf / resistance
= .1 / 4
= .025 A
= 25 mA.