Answer:
The minimum speed must the car must be 13.13 m/s.
Explanation:
The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.
We know that, mg be the weight of car and rider, which is equal to the centripetal force.
So, the minimum speed must the car must be 13.13 m/s.
Answer:
Follows are the solution to the given question:
Explanation:
In this question the missing file of the circuit is not be which is defined in attached file please find it.
In Option 1, this statement is true because the current is on , that is the same.
In option 2, this statement is false because therefore it implies that Rcd is always larger then
.
In option 3, this statement is true because the voltage of is always equal.
In option 4, this statement is true because is always smaller then 1 therefore,
is always equal to R1.
Answer:
The tangential speed at Livermore is approximately 284.001 meters per second.
Explanation:
Let suppose that the Earth rotates at constant speed, the tangential speed (), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:
(1)
Where:
- Rotation time, measured in seconds.
- Radius of the Earth, measured in meters.
- Latitude of the city above the Equator, measured in sexagesimal degrees.
If we know that ,
and
, then the tangential speed at Livermore is:
The tangential speed at Livermore is approximately 284.001 meters per second.