Answer:
Mass of star is 
kg.
Explanation:
The cube of orbital radius is equal to the square of its orbital time period is known as Kepler's law.
.....(1)
Here T is time period, r is orbital radius, G is universal gravitational constant and M is the mass of the star.
According to the problem,
Time period, T = 109 days = 109 x 24 x 60 x 60 s = 9.41 x 10⁶ s
Orbital radius, r = 18 AU = 18 x 1.496 x 10¹¹ m = 2.70 x 10¹² m
Gravitational constant, G = 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻²
Substitute these values in equation (1).
M = kg
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
The rock would be at a point 12 m from water at a time <u>4.8 s</u>.
Take the origin of the coordinate system at the top of the cliff. It is thrown upwards with a velocity u. When the rock is at a point 12 m from water, calculate the vertical displacement of the rock from the origin.
Use the equation of motion,
The rock falls under the acceleration due to gravity, directed down wards.
Substitute 18 m/s for u, -26 m for y and -9.8 m/s² for a=g.
Solve the quadratic equation for t.
Taking only the positive value,
After a time of <u>4.8 s</u> the rock would be at a distance of 12 m from water.
Answer:
11.962337 × 10^-4 N
Explanation:
Given the following :
Length L = 11.8
Charge = 29nC = 29 × 10^-9 C
Linear charge density λ = 1.4 × 10^-7 C/m
Radius (r) = 2cm = 2/100 = 0.02 m
Using the relation:
E = 2kλ/r ; F =qE
F = 2kλq/L × ∫dr/r
F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))
2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4
In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214
Hence,
(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N
