In calculating the energy of a photon of light, we need the relationship for energy and the frequency which is expressed as:
E=hv
where h is the Planck's constant (6.626 x 10-34 J s)and v is the frequency.
E = 6.626 x 10-34 J s (<span>7.33 x 10^14 /s) = 4.857 x 10^-19 J</span>
Answer:
When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
I HOPE ITS RIGHT
None of the choices is correct.
If two runners take the same amount of time to run a mile,
they have the same average speed. But their velocities
are not the same unless both runners begin and end their
run at the same points.
Speed is (distance covered) divided by (time to cover the distance).
Velocity is not. It's something different.
'Velocity' is not just a bigger word for 'speed'.
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
