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3 years ago

Figure 1 shows a wave movement during one second. What is the frequency of the wave

Physics

2 answers:

AURORKA [14]3 years ago

8 0

2 hertz, as explained previously- I'm afraid m/s is a completely different measurement

Andreyy893 years ago

4 0

This is 2 hertz. You can mark out 2 full wavelengths in the second of time.

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1 year ago

Ex 10: My dog runs at 6 m/s for 18 meters. How long did she run for?

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She ran for 3s

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2 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

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

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

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3 years ago

Help with physics projectile motion

BlackZzzverrR [31]

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

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3 years ago