<h2>
Answer: 56.718 min</h2>
Explanation:
According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
</em>
In other words, this law states a relation between the orbital period of a body (moon, planet, satellite) orbiting a greater body in space with the size of its orbit.
This Law is originally expressed as follows:
(1)
Where;
is the Gravitational Constant and its value is
is the mass of Mars
is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
(2)
(3)
(4)
Finally:
This is the orbital period of a spacecraft in a low orbit near the surface of mars
Answer:
2.49 * 10^(-4) m
Explanation:
Parameters given:
Frequency, f = 4.257 MHz = 4.257 * 10^6 Hz
Speed of sound in the body, v = 1.06 km/ = 1060 m/s
The speed of a wave is given as the product of its wavelength and frequency:
v = λf
Where λ = wavelength
This implies that:
λ = v/f
λ = (1060) / (4.257 * 10^6)
λ = 2.49 * 10^(-4) m
The wavelength of the sound in the body is 2.49 * 10^(-4) m.
Answer:
The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Explanation:
Given that
q₁ = 5 μ C
q₂ = - 4 μ C
The distance between charges = 50 cm
d= 50 cm
Lets take at distance x from the charge μ C ,the electrical field is zero.
That is why the distance from the charge - 4 μ C = 50 - x cm
We know that ,electric field is given as
Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.