The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;
Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
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The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery. Electrons would actually move through the wires in the opposite direction.
Since for closed system moles of the gas is always conserved
so as per idea gas equation we can say
so here we can say
so here we have
as we know that
now from above equation
on solving above equation we have
so here pressure will be 0.78 atm
Answer:
The correct answer is "164.06 nm".
Explanation:
Refractive index,
n = 1.75
Wavelength,
λ = 525 nm
Now,
The maximum thickness of the film will be:
⇒
On substituting the values, we get
⇒
⇒
⇒
The car's average acceleration would be 1.25m/s^2 or 1.25meters/second/second. That looks to be the fourth one you've listed.