How many orders of magnitude (powers of ten) is the distance to the moon greater than the distance across the Atlantic ocean?
1 answer:
Answer:
The distance to the moon greater than the distance across the Atlantic ocean by an order of 1.
Explanation:
We have the average distance across the Atlantic ocean,
the average distance to the moon,
The factor by which the distance to moon is greater than the width of Atlantic ocean:
i.e.
the distance to the moon greater than the distance across the Atlantic ocean by an order of 1.
You might be interested in
Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N
Answer:
Explanation:
Given data
To find
Mutual inductance of the two-coil system
Solution
The mutual inductance given as:
M= (-VΔt)/ΔI
Substitute the given values
So
Answer:
a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.
Explanation:
Statement is incomplete. The complete description is now described below:
<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>
<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>
<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>
Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (), measured in meters per second, is:
Where:
- Initial velocity in +x direction, measured in meters per second.
- Acceleration in +x direction, measured in meter per square second.
- Time, measured in seconds.
- Velocity in +y direction, measured in meters per second.
If we know that ,
,
and
, the final velocity of the satellite is:
a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:
The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:
The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.