So power is equal to work over time and work is force times distance, you do 5 times 3 and get 15 dividing by 2 gives us 7.5 W answer c
Answer:
q = 8.61 10⁻¹¹ m
charge does not depend on the distance between the two ships.
it is a very small charge value so it should be easy to create in each one
Explanation:
In this exercise we have two forces in balance: the electric force and the gravitational force
F_e -F_g = 0
F_e = F_g
Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.
Let's write Coulomb's law and gravitational attraction
In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.
k q² = G m²
q =
m
we substitute
q =
m
q =
m
q = 0.861 10⁻¹⁰ m
q = 8.61 10⁻¹¹ m
This amount of charge does not depend on the distance between the two ships.
It is also proportional to the mass of the ships with the proportionality factor found.
Suppose the ships have a mass of m = 1000 kg, let's find the cargo
q = 8.61 10⁻¹¹ 10³
q = 8.61 10⁻⁸ C
this is a very small charge value so it should be easy to create in each one
Answer:
∴ fractional compression = 1.34 × 10⁻²
Explanation:
given,
depth of Indian ocean = 3000 m
Bulk modulus of the water = 2.2 x 10⁹ N/m²
We know,
P = P₀ + ρgh
P₀ is the atmospheric pressure
P₀ = 10⁵ N/m²
ρ is the density of the water, 1000 Kg/m³
P = 10⁵ + 1000 × 9.8 × 3000 = 2.94 × 10⁷ N/m²
using formula,
B = P/{-∆V/V}
B is bulk modulus and { -∆V/V} is the fractional compression


∴ fractional compression = 1.34 × 10⁻²