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3 years ago

If there are 20 waves in a 5-second time frame, what is the frequency?

Physics

2 answers:

mestny [16]3 years ago

6 0

Answer:

The frequency is four waves per second.

Explanation:

You make it so the time is one second.

To do that you divide 5 seconds by 5.

Since you divide the seconds by 5 you now have to divide the amount of waves by 5.

20/5=4 5/5=1

Now your answer is four waves per second.

creativ13 [48]3 years ago

3 0

Your answer would be 0.20 Hz

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The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40cm. the lift raises a lo

Elena L [17]

Sm = small piston
la = large piston

P=F/A
P=15000/(20^2)π

F of sm = PA
= (75/2π)•((8^2)π)
= (75•64)/2
= 4800/2
= 2400N
We already know the pressure but giving it in an approximate decimal form, to two significant figures (since that's what your supplied precision is at):

a) 12 Pa
b) 2400 N

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3 years ago

Read 2 more answers

At a maximum level of loudness, the power output of a 75-piece orchestra radiated as sound is 72.0 W. What is the intensity of t

FrozenT [24]

Answer:

The intensity would be 0.01432 W/ $m^{2}$

Explanation:

The sound intensity is the power that is transmitted by a sound wave per unit area.

Given that

Power output P of the 75-piece orchestra = 72.0 W

The listener's distance r = 20.0 m

The intensity of sound waves (I) can be obtained with the expression below;

I = P/A .............. 1

where P is the power

A is the area, in this case, at the distance r the sound is radiated through a sphere.

A = area of a sphere = 4π $r^{2}$

putting it into equation 1, the intensity would be;

I = P/ 4π $r^{2}$

Substituting our values we have;

I = 72.0 W / 4 π $20^{2}$ $m^{2}$

I = 0.01432 W/ $m^{2}$

Therefore the intensity would be 0.01432 W/ $m^{2}$

3 0

3 years ago

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³

Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, $\rho_o$ = 800 kg/m³

density of water, $\rho_w$ = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

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3 years ago

Though you add heat, the temperature of boiling water remains constant because ...

Harlamova29_29 [7]

The temperature rises during boiling.

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3 years ago

The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.

Bad White [126]

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

The angular velocity of the telescope is approximately 0.199 rad/s

Second question:

The telescope should accelerates away by approximately F = 0.0005·m

Third question:

The pulling force between the Earth and the satellite

<h3>What equations can be used to calculate the velocity and forces acting on the telescope?</h3>

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

$Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}$

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

$Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199$

The angular velocity of the telescope, ω ≈ 0.199 rad/s

Second question:

Centrifugal force force, $F_{\omega}$ = m·ω²·r

Which gives;

$F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m$

$Gravitational \ force, F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}$

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

$F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx 0.00592 \cdot m$

Which gives;

$F_{\omega}$ < $F_G$ , therefore, the James Webb telescope has to accelerate away from the Sun

F = $\mathbf{F_{\omega}}$ - $\mathbf{F_G}$

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ 0.0005·m (away from the Sun)

Third part:

Other forces include;

The force of attraction between the Earth and the telescope which can contribute to the the telescope having a stable orbit at the given speed.

Learn more about orbital motion here:

brainly.com/question/11069817

3 0

3 years ago