A tiger leaps horizontally froma 7.5 m high rock with a speed of 3.0 m/s. How far from the base of the rock will she land?
1 answer:
Answer:
<em>The tiger will land at 3.71 meters from the base of the rock</em>
Explanation:
<u>Horizontal Motion</u>
When an object is thrown horizontally with a speed vo from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:
The tiger leaps from a height of h=7.5 m with a speed of v=3 m/s. Substituting the values into the formula:
The tiger will land at 3.71 meters from the base of the rock
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Answer:
λ = 5.85 x 10⁻⁷ m = 585 nm
f = 5.13 x 10¹⁴ Hz
Explanation:
We will use Young's Double Slit Experiment's Formula here:
where,
λ = wavelength = ?
Y = Fringe Spacing = 6.5 cm = 0.065 m
d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m
L = screen distance = 5 m
Therefore,
<u>λ = 5.85 x 10⁻⁷ m = 585 nm</u>
Now, the frequency can be given as:
where,
f = frequency = ?
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>f = 5.13 x 10¹⁴ Hz</u>