The answer is Prove a hypothesis
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
A) From rest and before performing the described exercise, the entire weight of the desk is supported by the floor. This support force (normal force) exerted by the floor will be equivalent to the force equivalent to the weight of the desk. From the moment the desk starts to be lifted, said support force will decrease and the weight of the desk supported by the floor will be less.
B) As the desk is lifted, the support force of the two guests will increase, this occurs because the force is transferred to the support of both, but the support force of the floor on the two people will increase.
In triangle ABC
Sin5 = BC/AC
BC = (AC) Sin5
BC = (50) Sin5 = 4.36 cm
r = distance between the two balls = 2 BC = 2 x 4.36 = 8.72 cm = 0.0872 m
q = charge on each ball
m = mass of each ball = 5 g = 0.005 kg
electric force between the two balls is given as
F =
using equilibrium of force in vertical direction
T Cos5 = mg eq-2
Using equilibrium of force in horizontal direction
T Sin5 = F eq-3
dividing eq-3 by eq-2
T Sin5 /(T Cos5) = F/mg
F = mg tan5
using eq-1
= mg tan5
inserting the values
= 0.005 x 9.8
q = 2.03 x 10⁻⁷ C
sign of charge is same on both the balls. either it is negative or positive