The gravitational force between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the first object has a mass of
, while the second "object" is the Earth, with mass
. The distance of the object from the Earth's center is
; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
Answer:
ρ/ρ2 = 3 / R₀ the two densities are different
Explanation:
Density is defined as
ρ = M / V
As the nucleus is spherical
V = 4/3 π r³
Let's replace
ρ = A / (4/3 π R₀³)
ρ = ¾ A / π R₀³
b)
ρ2 = F / area
The area of a sphere is
A = 4π R₀²
ρ2 = F / 4π R₀²
ρ2 = F / 4π R₀²
Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.
Let's look for the relationship of the two densities
ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)
ρ /ρ2 = 3 (A / F) (1 / R₀)
In this case it does not say that the nucleon number is A (F = A), the relationship is
ρ/ρ2 = 3 / R₀
I see that the two densities are different
Answer:
B) 12 m
Explanation:
Gravitational potential energy is:
PE = mgh
Given PE = 5997.6 J, and m = 51 kg:
5997.6 J = (51 kg) (9.8 m/s²) h
h = 12 m
Answer:
Final velocity of electron,
Explanation:
It is given that,
Electric field, E = 1.55 N/C
Initial velocity at point A,
We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :
........(1)
a is the acceleration,
We know that electric force, F = qE
Use above equation in equation (1) as:
v = 647302.09 m/s
or
So, the final velocity of the electron when it reaches point B is . Hence, this is the required solution.
Answer:
The first law states that if the net force is zero, then the velocity of the object is constant.