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4 years ago

An echo repeats two syllables.If the velocity of sound is 330 m/s , then the distance of the reflecting surface is-

1 answer:

6 0

Well, the main point you should remember is that we only can hear 20hz that leads to T=0.05sec which means that the distance leads us to velocity x time.
When calculating, you get 330 * 0.05=16.5 .
The number which you got from the previous step has to me multiplied 4 times (4 minutes) and you will get the answer you need - a) 66.0 m.
Do hope it will help you! Regards!

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Sound waves cannot carry energy through. A water B air C a mirror D a vacuum

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I looked up the question and got D- a vacuum

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3 years ago

How will the amount of power change if less work is done in more time?

yKpoI14uk [10]

The amount of power change if less work is done in more time"then the amount of power will decrease".

Option: B

Explanation:

The rate of performing any work or activity by transferring amount of energy per unit time is understood as power. The unit of power is watt

$Power = \frac{Work}{Time}$

Here this equation showcase that power is directly proportional to the work but dependent upon time as time is inversely proportional to the power i.e as time increases power decreases and vice versa.

This can be understood from an instance, on moving a load up a flight of stairs, the similar amount of work is done, no matter how heavy but when the work is done in a shorter period of time more power is required.

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3 years ago

A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surf

liberstina [14]

Answer:

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.

Explanation:

According to the following equation:

$\frac{n_{w} }{s} +\frac{n_{a} }{s'} = \frac{n_{a}- n_{w}}{R_{c} } \\$

where nw and na is the refractive indices of water (1.33) and air (1.00) respectively; s is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.

Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).

Therefore, the above equation can be expressed as:

$\frac{n_{w}}{s} +\frac{n_{a}}{s'}=0$

Now we can solve for the apparent depth of the fish.

(a) $s'=-(\frac{n_{a}}{n_{w}})x s$ (Make s' subject of the formula from the above equation)

$s'=(\frac{1.00}{1.33} )x7cm$

∴ $s'=5.3 cm.$

(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.

Therefore, s = 33 cm

$s'=-(\frac{n_{a}}{n_{w}})x s$

$s'=(\frac{1.00}{1.33} )x 33 cm$

$s'=24.8 cm.$

NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

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4 years ago

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kvasek [131]

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sattari [20]

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