Answer:
Explanation:
The relation between time period of moon in the orbit around a planet can be given by the following relation .
T² = 4 π² R³ / GM
G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .
Substituting the values in the equation
(.3189 x 24 x 60 x 60 s)² = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)
759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )
M = .06424 x 10²⁵
= 6.4 x 10²³ kg .
Answer:
t = 12,105.96 sec
Explanation:
Given data:
weight of spacecraft is 2000 kg
circular orbit distance to saturn = 180 km
specific impulse = 300 sec
saturn orbit around the sun R_2 = 1.43 *10^9 km
earth orbit around the sun R_1= 149.6 * 10^ 6 km
time required for the mission is given as t
![t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%5Cmu_sun%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28R_1%20%2B%20R_2%29%5D%5E%7B3%2F2%7D)
where
is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2.![t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%201.32712%20x%2010%5E%7B20%7D%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28149.6%20%2A%2010%5E%206%20%2B1.43%20%2A10%5E9%20%29%5D%5E%7B3%2F2%7D)
t = 12,105.96 sec
Answer:
Electromagnetic waves consist of both electric and magnetic field waves. These waves oscillate in perpendicular planes with respect to each other, and are in phase. The creation of all electromagnetic waves begins with an oscillating charged particle, which creates oscillating electric and magnetic fields.
Explanation:
Answer:
3.8 x 10^10 m/s^2
Explanation:
Charge, q = 2 e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C
Electric field strength, E = 790 N/C
mass of helium nucleus, m = 6.645 x 10^-27 Kg
the force due to electric filed on a charge particle is given by
F = q x E
Where, q be the charge on the charged particle and E be the strength of electric field.
By substituting the values
F = 3.2 x 10^-19 x 790
F = 2528 x 10^-19 N
According to the Newton's second law
F = m x a
Where, me be the mass and a be the acceleration
By substituting the values
a = 3.8 x 10^10 m/s^2
The precision of the measuring instruments will be 2.5%.
<h3>What is precision?</h3>
It should be noted that precision simply means how close a test result will be when it's repeated.
In this case, the precision of the measuring instruments will be:
= (2 - 1.95)/2.
= 0.05/2
= 2.5%
Therefore, the the precision of the measuring instruments will be 2.5%.
Learn more about precision on:
brainly.com/question/5792909
