Answer:
The magnitude of gravitational force between two masses is
.
Explanation:
Given that,
Mass of first lead ball, 
Mass of the other lead ball, 
The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m
We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

So, the magnitude of gravitational force between two masses is
. Hence, this is the required solution.
Answer:
(a) - 42700 m/s
(b) - 6.8 x 10^-4 m/s^2
Explanation:
initial velocity of star, u = 20.7 km/s
Final velocity of star, v = - 22 km/s
time, t = 1.99 years
Convert velocities into m/s and time into second
So, u = 20700 m / s
v = - 22000 m/s
t = 1.99 x 365.25 x 24 x 3600 = 62799624 second
(a) Change in planet's velocity = final velocity - initial velocity
= - 22000 - 20700 = - 42700 m/s
(b) Accelerate is defined as the rate of change of velocity.
Acceleration = change in velocity / time
= ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I
Gravitational force = G ( m1 m2 ) / r²
3 = G ( m1 m2 ) / ( 10 )²x = G ( m1 m2 ) / ( 5 )²We shall divide those two equations:3 / x = 1/100 / 1/25 = 25 / 100 = 1 / 4x · 1 = 3 · 4x = 12Answer:C. 12 N