Answer: V = 15 m/s
Explanation:
As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,
F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz
Using doppler effect formula
F = C/ ( C - V) × f
Where
F = observed frequency
f = source frequency
C = speed of light = 3×10^8
V = speed of the car
Substitute all the parameters into the formula
2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10
2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)
1.000000049 = 3×10^8/(3×10^8 - V)
Cross multiply
300000014.7 - 1.000000049V = 3×10^8
Collect the like terms
1.000000049V = 14.71429
Make V the subject of formula
V = 14.71429/1.000000049
V = 14.7 m/s
The speed of the car is 15 m/s approximately
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-Slow down
-Speed up
-Turning
Hope this helps
the answer is foot luv <3
Answer:
Approximately 21 km.
Explanation:
Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:
- No intersection: There's nothing that blocks the camera's view of the top of the building.
- Two intersections: The planet blocks the camera's view of the top of the building.
- One intersection: The point at which the top of the building appears or disappears.
There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.
The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle
which corresponds to this minor arc.
This angle comes can be split into two parts:
.
Also,
.
The radius of this circle is:
.
The lengths of segment DC, AC, BC can all be found:
In the two right triangles
and
, the value of
and
can be found using the inverse cosine function:


.
The length of the minor arc will be:
.