A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
1 answer:
Answer:
The maximum height reached by the body is 313.6 m
The time to return to its point of projection is 8 s.
Explanation:
Given;
initial velocity of the body, u = 78.4 m/s
at maximum height (h) the final velocity of the body (v) = 0
The following equation is applied to determine the maximum height reached by the body;
v² = u² - 2gh
0 = u² - 2gh
2gh = u²
h = u²/2g
h = (78.4²) / (2 x 9.8)
h = 313.6 m
The time to return to its point of projection is calculated as follows;
at maximum height, the final velocity becomes the initial velocity = 0
h = v + ¹/₂gt²
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
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Answer:
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Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
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we substitute
h f = K + Ф
Ф= hf - K
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let's reduce the energy to the SI system
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calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
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Ф = 2.179 eV