That the radio wave has more to it
The relationship between the impulse and the momentum allows to find the result for the question about the contact time is:
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The contact time on the floor is t = 0.60 s
The impulse is the push force on a body and is related to the variation of the amount of momentum.
I = ∫ F. dt = Δp
p = mv
where I is the momentum, F the force, t the time, p the amount of momentum and m the mass .
They indicate the average force F = 330 N, the jump height is y = 12.0 cm = 0.120 m.
F. t = m v_f - m v₀
Let's search with kinematics for the initial speed of the jump.
v² = v₀² - 2 g y
At the highest point the velocity is zero.
0 = v₀² - 2ay
Let's calculate.
v₀ = 1.53 m / s
Suppose that the jumps are elastic, that is, that the speed with which it reaches the ground is equal to the speed with which it leaves, but in the opposite direction.
Let's calculate.
t = 0.60 s
In conclusion using the relationship between impulse and momentum we can find the result for the question about the contact time is:
The contact time on the floor is t = 0.60 s
Learn more about impulse here: brainly.com/question/904448
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Answer:
the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m
Explanation:
Given the data in the question;
slit separation d = 0.165 mm = 0.165 × 10⁻³ m
wavelength λ = 560 nm = 560 × 10⁻⁹ m
distance between the screen and slits D = 4.05 m
now,
for fifth-order bright fringe path difference = mλ
where m is 5
so, the difference in path lengths from each of the slits will be;
Δr = mλ
we substitute
Δr = 5( 560 × 10⁻⁹ m )
Δr = 28 × 10⁻⁷ m
Therefore, the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m
Answer: energy source, path, and load
Explanation: