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Ainat [17]
4 years ago
11

A submarine is 58.8 m from a whale. The sub sends out a sonar ping to locate the whale. The speed of sound underwater is 1520 m/

s. How much time does it take for the sound wave to travel to the whale and back? (Unit = s)
Physics
1 answer:
MissTica4 years ago
6 0

Answer:

0.08

Explanation:

this problem assume that both of whale and submarine are in rest position or in constant linier motion in same direction and same speed.

The sound will travel from Submarine to the whale and back again to submarine. so the time will be like this

t = 2d/v

t = 2*58.8/1520

t = 117.6/1520

t = 0.077368 s

t ≈ 0.08 s (less then 1 s)

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Answer:

f.The period is independent of the suspended mass.

Explanation:

The period of a pendulum is given by

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where

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g is the acceleration due to gravity

From the formula, we see that:

1) the period of the pendulum depends only on its length, L, and it is proportional to the square root of the length

2) the period does not depend neither on the mass of the pendulum, nor on its amplitude of oscillation

So, the only correct statements are

f.The period is independent of the suspended mass.

Note: statement "e.The period is proportional to the length of the wire" is also wrong, because the period is NOT proportional to the length of the wire, but it is proportional to the square root of it.

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The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah
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The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.1
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Answer:

Minimum number of photons required is 1.35 x 10⁵

Explanation:

Given:

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Energy of one photon is given by the relation :

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Here h is Planck's constant and c is speed of light.

Let N be the minimum number of photons needed for triggering receptor.

Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J

According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,

E₁ = N x E

Put equation (1) in the above equation.

E_{1}=N\times\frac{hc}{\lambda}

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3.15\times10^{-14} =N\times\frac{6.6\times10^{-34}\times3\times10^{8}}{850\times10^{-9}}

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Answer:

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