Answer:
5.23km/s
Explanation:
Given
Radius of Earth = 6.37 * 10^6 m
Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m
Gravity Acceleration on Satellite Altitude = 1.87965m/s²
For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.
Centripetal Acceleration = V²/R
So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R
Make V the subject of formula
A = V²/R
V² = AR
V = √AR
Where R = (radius of earth) + (altitude of satellite)
R = 6.37 * 10^6 + 8.2 * 10^6
R = 14.57 * 10^6m
A = 1.87965m/s²
V = √(1.87965 * 14.57x10^6)
V = √27386500.5
V = 5233.211299001789
V = 5233.2113 m/s ------- Approximated
V = 5.23km/s
Answer:
2.7km
Explanation:
Two methods: Convert km/hour to km/minutes or convert 3/2 minutes to hours.
Then multiply time to get the distance of the car traveled.
Given mass= 1kg
Weight on earth = mg(gravity of earth) = 9.8N
weight on moon = mg(gravity of moon)= 1.62N
weight on outer space mg(gravity outer space = 0) = 0N
Answer:
A. The closest point in the Moon's orbit to Earth
Explanation:
The perigee is defined as the closest point in the orbit of an object (such as a satellite) from the centre of the Earth. In this case, the Earth's satellite is the Moon, so the perigee is defined as the closest point in the Moon's orbit to Earth. so option A is the correct one.
Let's see instead the names of the other options:
B. The farthest point in the Moon's orbit to Earth --> this point is called apogee
C. The closest point in Earth's orbit of the Sun --> this point is called perihelion
D. The Sun's orbit that is closest to the Moon --> this point has no specific name
Answer:
35.14°C
Explanation:
The equation for linear thermal expansion is 
, which means that a bar of length 
with a thermal expansion coefficient 
under a temperature variation 
will experiment a length variation 
.
We have then = 0.481 foot, = 1671 feet and = 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a 
= 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.
Since our temperature on a cool spring day was 13.0°C, our new temperature must be 
= 35.14°C
