Answer:
252.45 hours or 908820 seconds
Explanation:
The equatorial radius of Saturn is 60,268 km
The length of the equator will be circumference
Speed of the equatorial flow = 1500 km/h
It will take 252.45 hours or 908820 seconds for the equatorial flow to encircle the planet.
Answer:
The wife have to sit at 0.46 L from the middle point of the seesaw.
Explanation:
We need to make a sketch of the seesaw and the loads acting over it.
And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.
A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.
The maximum momentum with respect to the pivot point (0) will be:
Where L/2 is the half of the distance of the seesaw
Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.
Answer:
density is Mg/µL
Explanation:
given data
density of nuclear = kg/m³
1 ml = 1 cm³
to find out
density of nuclear matter in Mg/µL
solution
we know here
1 Mg = 1000 kg
so
1 m³ is equal to cm³
and here 1 cm³ is equal to 1 mL
so we can say 1 mL is equal to 10³ µL
so by these we can convert density
density = kg/m³
density = kg/m³ ×
Mg/µL
density = Mg/µL
Answer:
Explanation:
Given:
<u>The relation between the change in pressure with height is given as:</u>
where:
dz = height in the atmosphere
= standard value of gravity
<em>Now putting the respective values:</em>
Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)
Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
