Answer:
r = 4.21 10⁷ m
Explanation:
Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining
T² = (
) r³ (1)
in this case the period of the season is
T₁ = 93 min (60 s / 1 min) = 5580 s
r₁ = 410 + 6370 = 6780 km
r₁ = 6.780 10⁶ m
for the satellite
T₂ = 24 h (3600 s / 1h) = 86 400 s
if we substitute in equation 1
T² = K r³
K = T₁²/r₁³
K =
K = 9.99 10⁻¹⁴ s² / m³
we can replace the satellite values
r³ = T² / K
r³ = 86400² / 9.99 10⁻¹⁴
r = ∛(7.4724 10²²)
r = 4.21 10⁷ m
this distance is from the center of the earth
Answer:
The magnitude of the resultant acceleration is 2.2 
Explanation:
Mass (m) of the sailboat = 2000 kg
Force acting on the sailboat due to ocean tide is
= 3000N
Eastwards means takes place along the positive x direction
Then
= 3000N and
= 0
Wind Force acting on the Sailboat is
= 6000N directed towards the northwest that means at an angle 45 degree above the negative x axis
Then
= -(6000N) cos 45 degree = -4242.6 N
= (6000N) cos 45 degree = 4242.6 N
Hence , the net force acting on the sailboat in x direction is

= - 3000 N + 4242.6 N
= - 3000 N +4242.6 N
= 1242.6N
Net Force acting on the sailboat in y direction is

= 0+ 4242.6N
= 4242.6N
The magnitude of the resultant force =
Using pythagorean theorm of 1243 N and 4243 N



4420.8 N
F = ma


=2.2 
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
Answer with Explanation:
We are given that


Charge on proton,q=
a.We have to find the electric potential of the proton at the position of the electron.
We know that the electric potential

Where 


B.Potential energy of electron,U=
Where
Charge on electron
=Charge on proton
Using the formula


km x h = km/h
First trial: 6 x 1 = 6km/h
Second trial: 9 x 2 = 18km/h
6 + 18 = <u>24km/h</u> (Total)
Or
6 + 9 = 15 km
2 + 1 = 3h
15 + 3 = 18
15 x 2 = 30
3 x 2 = 6
30 - 6 = <u>24km/h</u>