speed of tortoise is given as v1 = 0.14 m/s
speed of hare is given as v2 = 20*0.14 = 2.8 m/s
now let say the total length of the path is "d"
so the total time taken by the tortoise to cover this
now given that hare took rest for 1 min
so total time of run for hare is (t - 60)s
so the distance that hare covered is given by
now by above two equations
and the time t is given by
so part a)
t = 63 s
part b)
d = 8.82 m
Answer:
θ = 10.28º
Explanation:
To find the angle of refraction use the equation of refraction
n₁ sin θ₁ = n₂ sin θ₂
where index 1 is for incident light and index 2 is for refracted light.
sin θ₂ = n₁ / n₂ sin θ
let's calculate
sin = 1 / 1.3 sin 0.23
sin = 0.175
θ= 0.17528 rad
let's reduce to degrees
θ = 0.17528 rad (180ª / pi rad)
θ = 10.28º
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.
D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .
-- They don't change by the same factor, because 1/g is inside
the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value
of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
Answer:
(A)
Explanation:
We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.
Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.
