2. An object of mass 20 kg is lifted to a 25 m building. How much potential energy is stored on the mass? (Take g= 10 m/s)
1 answer:
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Answer:
486nm
Explanation:
in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that
Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state
Answer:
1/3 the distance from the fulcrum
Explanation:
On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:
where
W1 is the weight of the boy
d1 is its distance from the fulcrum
W2 is the weight of his partner
d2 is the distance of the partner from the fulcrum
In this problem, we know that the boy is three times as heavy as his partner, so
If we substitute this into the equation, we find:
and by simplifying:
which means that the boy sits at 1/3 the distance from the fulcrum.
Answer:
a) v₀ = 32.64 m / s , b) x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
x = v₀ₓ t
y = t - ½ g t²
We look for the components of speed with trigonometry
sin 43 = v_{oy} / v₀
cos 43 = v₀ₓ / v₀
v_{oy} = v₀ sin 43
v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
t = x / v₀ cos 43
y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
y = x tan 43 - ½ g x² / v₀² cos² 43
1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
v₀ = √ (35280 / 33.11)
v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
y = t - ½ g t²
.y = v₀ sin43 t - ½ g t²
25 = 32.64 sin 43 t - ½ 9.8 t²
4.9 t² - 22.26 t + 25 = 0
t² - 4.54 t + 5.10 = 0
We solve the second degree equation
t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
t = (4.54 ± 0.46) / 2
t₁ = 2.50 s
t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
x = v₀ₓ t
x = v₀ cos 43 t
x = 32.64 cos 43 2.50
x = 59.68 m