The object is farther away then two principles focal lengths from the concave lens
Answer:
Option (1), option (4) and option (5)
Explanation:
The main observations of Ernest Rutherford's experiment are given below:
1. most of the positively charged particles pass straight, it means there is an empty space in the atom.
2. Very few positively charged particles retraces their path.
So,
The positively charged particles were deflected because like charges repel, that means they are deflected by protons.
Almost all the positively charge concentrate in a very small part which is called nucleus.
Answer:
M = 1.38 10⁵⁹ kg
Explanation:
For this problem we will use the law of universal gravitation
F = G m₁ m₂ / r²
Where G is the gravitation constant you are value 6.67 10⁻¹¹ N m2 / kg2, m are the masses and r the distance
In this case the mass of the planet is m = 3.0 10²³ kg and the mass of the start is M
Let's write Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
The speed module is constant, so we can use the kinematic relationship
v = d / t
The distance remembered is the length of the circular orbit and the time in this case is called the period
d = 2π r
a = 2π r / T
Let's replace Newton's second law
G m M / r² = m (4π² r² / T²) / r
G M = 4 π² r³ / T²
M = 4 π² r³ / T² G
Let's calculate
M = 4 π² (3.0 10²³)³ / (3.4 10¹¹)² 6.67 10⁻¹¹
M = 13.82 10⁵⁸ kg
M = 1.38 10⁵⁹ kg
Answer:
I_total = L² (m + M / 3)
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
It is appreciated that it is a scalar quantity, for which it is additive, in this case the system is formed by two bodies and the moment of inertia must be the sum of each moment of inertia with respect to the same axis of rotation.
The moment of inertia of a bar with respect to an axis that passes through ends is
I_bar = 1/3 M L²
The moment of inertia of a particle is
I_part = m x²
We have to assume the point where the particle sticks to the bar, suppose it sticks to the end
x = L
Total moment of inertia is the sum of these two moments of inertia
I_total = I_bar + I_particule
I_total = 1/3 M L² + m L²
I_total = L² (m + M / 3)