Answer:
6 m/s
Explanation:
To determine the change in speed of the object, we just need to determine its speed at t = 30 s and at t = 20 s, and then calculate the difference.
The speed at t = 30 is:
v = 6 m/s
While the speed at t = 20 s is:
u = 0
Therefore, the change in speed is:
Sometimes, because acceleration due to gravity on Earth depends on how close you are to the Earth's center.
Answer:
T= 37 day
Explanation:
To solve this exercise we will use the definition of angular velocity as the angular distance, which for a full period is 2pi between time.
w = T / t
The relationship between angular and linear velocity is
v = w r
w = v / r
We substitute everything in the first equation
v / r = 2π / t
t = 2π r / v
Let's reduce to the SI system
V = 2 km / s (1000m / 1km) = 2 10³ m / s
r= R = 6.96 10⁸ m
Let's calculate
t = 2π 6.96 10⁸/2 10³
t = 3.2 10⁶ s
T = t = 3.2 10⁶ s ( 1h/3600s) (1 day/24 h)
T= 37 day
<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
