A ball is thrown upwards with an initial velocity of 25.o m/s, what is the velocity of the ball at 11.9 m from the ground?
1 answer:
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First, illustrate the problem as shown in the attached picture. Next, let's find the distance traveled by planes A and B after 2.9 h.
Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m
Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°
x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>
Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m
Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°
x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>
Answer:
3.71 eV
Explanation:
λ = Wavelength of light = 224 nm = 224 x 10⁻⁹ m
c = speed of electromagnetic wave = 3 x 10⁸ m/s
V₀ = stopping potential = 1.84 volts
W₀ = Work function of the metal = ?
Using the equation
= 5.94 x 10⁻¹⁹
= 3.71 eV