Answer:
it's important because it shows how thermal energy transforms or continues to be all around us in everything
Answer:
Explanation:
Let i be the angle of incidence and r be the angle of refraction .
From the figure
Tan ( 90 - i ) = 2.5 / 8
cot i = 2.5 / 8
Tan i = 8 / 2.5 = 3.2
i = 72.65°
From snell's law
sini / sin r = refractive index
sin 72.65 / sinr = 1.333
sin r = .9545 / 1.333
= .72
r = 46⁰
From the figure
Tan r = d / 4
Tan 46 = d /4
d = 4 x Tan 46
= 4 x 1.0355
=4.14 m .
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ =
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ =
F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
-- The car starts from rest, and goes 8 m/s faster every second.
-- After 30 seconds, it's going (30 x 8) = 240 m/s.
-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s
-- Distance covered in 30 sec at an average speed of 120 m/s
= <span> 3,600 meters .</span>
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The formula that has all of this in it is the formula for
distance covered when accelerating from rest:
Distance = (1/2) · (acceleration) · (time)²
= (1/2) · (8 m/s²) · (30 sec)²
= (4 m/s²) · (900 sec²)
= 3600 meters.
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When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.
When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.
How does he do that ?
By accelerating at 8 m/s². That's about 0.82 G !
He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !
He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion

Here s = h,u = 450m/s a = -g and t = t+3
Substituting

Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting

Solving both equations

So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s

Passing of B occurs at 4108.31 height.