The cliff is 319. kilograms above the water
Answer:
So the force of attraction between the two objects is 3.3365*10^-6
Explanation:
m1=10kg
m2=50kg
d=10cm=0.1m
G=6.673*10^-11Nm^2kg^2
We have to find the force of attraction between them
F=Gm1m2/d^2
F=6.673*10^-11*10*50/0.1^2
F=3.3365*10^-8/0.01
F=3.3365*10^-6
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:

Answer:
D = 527.31 Km
Explanation:
given,
angle of ship, θ = 23.5° N of W
distance travel in the direction = 575 Km
Distance of ship in west from harbor = ?
now,
Distance of the ship in the west direction
D = d cos θ
d = 575 Km
θ = 23.5°
inserting all the values
D = 575 x cos 23.5°
D = 575 x 0.91706
D = 527.31 Km
Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km
Its Displacement and Time for sure.
Thank You!
Pls mark Brainliest!