Answer:
12 km/h
Explanation:
Average Speed = Distance / Time (or rate)
Pick a point on the graph for Ian and plug in values.
For example, 20 minutes -> 4km
Hence, Average speed = 4km ÷ 20 minutes = 0.2 km/min
0.2 km/min × 60 = 12 km/h
If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be
.
<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be
.
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Answer:
So the distance of the antenna from the station will be 3.018 m
Explanation:
We have given the frequency of the broadcast 
The speed of light 
The distance of the antenna to receive a minimum signal from the station is given by 
So the distance of the antenna from the station will be 3.018 m
Answer:
Her speed is 1.1 m/s, and her velocity is 0 m/s
Explanation:
Speed = Distance covered/Time
Given
Distance = 400m
Time = 6minutes = 6*60 = 360 secs
Substitute the given parameter into the formula;
Speed = 400/360
Speed = 1.1m/s
Since the track is a circular track, the displacement will be zero. She is only moving in a circular path (no direction)
Velocity = Displacement/Time
Velocity = 0/3600
Velocity = 0m/s
Hence her speed is 1.1 m/s, and her velocity is 0 m/s
Answer:
![[\psi]= [Length^{-3/2}]](https://tex.z-dn.net/?f=%5B%5Cpsi%5D%3D%20%5BLength%5E%7B-3%2F2%7D%5D)
- This means that the integral of the square modulus over the space is dimensionless.
Explanation:
We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

must be dimensionless, as represents a probability.
As the differentials has units of length
for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:
![[\psi]^2 = [Length^{-3}]](https://tex.z-dn.net/?f=%5B%5Cpsi%5D%5E2%20%3D%20%5BLength%5E%7B-3%7D%5D)
taking the square root this gives us :
![[\psi] = [Length^{-3/2}]](https://tex.z-dn.net/?f=%5B%5Cpsi%5D%20%3D%20%5BLength%5E%7B-3%2F2%7D%5D)