Answer:
Sorry don't know the answer
Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
Answer:
Explanation:
position of centre of mass of door from surface of water
= 10 + 1.1 / 2
= 10.55 m
Pressure on centre of mass
atmospheric pressure + pressure due to water column
10 ⁵ + hdg
= 10⁵ + 10.55 x 1000 x 9.8
= 2.0339 x 10⁵ Pa
the net force acting on the door (normal to its surface)
= pressure at the centre x area of the door
= .9 x 1.1 x 2.0339 x 10⁵
= 2.01356 x 10⁵ N
pressure centre will be at 10.55 m below the surface.
When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .
Answer:
b
Explanation:
friction needs two objects
the two objects need to touch each other
the two objects need to either have relative movement or, tend to have relative movement urge. (kinetic friction or static friction)