3 years ago

Does anyone know what the answers are?

Physics

1 answer:

baherus [9]3 years ago

4 0

Answer:

Option (C) is the answer

Explanation:

may be it is possible if that we stand so far

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It’s light seconds, now don’t mistake it as time

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What is the lupac name H3C – CHT the H₂ - CH - CH2 CH₂ CH₂ CH3

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Answer:??

Explanation:

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3 years ago

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

$I_2 = 4W/m^2$

Therefore the intensity at five times this distance from the source is $4W/m^2$

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3 years ago

The acceleration due to gravity on the moon is 1.6 m/s2. What is the gravitational potential energy of a 1200-kg lander resting

faltersainse [42]

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The equation for gravitational potential energy is GPE = mgh.

GPE = 1200(1.6)(350) = 672000 J

Hope this answers the question. Have a nice day.

5 0

3 years ago

Read 2 more answers

A 4-N object object swings on the end of a string as a simple pendulum. At the bottom of the swing, the tension in the string is

Amanda [17]

Answer:

Explanation:

Given mg = 4N .

m = 4 / g

At the bottom of the swing let centripetal acceleration be a

T - mg = ma

9 - 4 = ma

5 = 4 a / g

a = 5g / 4

6 0

3 years ago