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3 years ago

If the moon disappeared what effect would this have on the earths tides

2 answers:

iogann1982 [59]3 years ago

4 0

The moon has a huge impact on the tides. With the Moon gone, the oceans would become much calmer. The Sun still has its effect on them known as the solar tides. Surfers wouldn't be completely devoid of waves.

Sliva [168]3 years ago

3 0

There would no longer be any tides.

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A slinky forms it’s third harmonic standing wave when the input frequency is 24 Hz. What is the fundamental frequency of the sli

jarptica [38.1K]

Answer: The fundamental frequency of the slinky = 8Hz

An input frequency of 28 Hz will not create a standing wave

Explanation:

Let Fo = fundamental frequency

At third harmonic,

F = 3Fo

If F = 24Hz

24 = 3Fo

Fo = 24/3 = 8Hz

If an input frequency = 28 Hz at 3rd harmonic

Let find the fundamental frequency

28 = 3Fo

Fo = 28/3

Fo = 9.33333Hz

Since Fo isn't a whole number, it can't create a standing wave

6 0

4 years ago

An object starts from rest and uniformly accelerates at a rate of 1.25 m/s2 for 7.0 seconds.

stich3 [128]

Explanation:

Since its accelerating, the velocity vs time graph is linear

For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula

A= (vf - vi)/t

a= vf-0/t

1.25=vf / 7

1.25*7=vf

8.75 = vf

Now for displacement plug all the values in

X = 1/2(vf-vi)/t formula

The displacement (x) is 30.625 m

For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above

The answer is t = 5.02

Im pretty sure all the answers are correct

8 0

2 years ago

Menciona un ejemplo de la vida cotidiana donde se apliquen las cuatro fuerzas fundamentales de la naturaleza.

Gekata [30.6K]

Yesterday was the day that we got canceled on it so we had a lot

4 0

3 years ago

If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m

Margaret [11]

Ideal Gas Law is, pV = NkbT

<span>Therefore, p/t = Nkb/V which is equal to the constant</span>

We need to convert the given temperature to Kelvin. We need to add 273 to have the Kelvin of the temperature from Celsius.

T1= 20 + 273 = 293 K

T2= 120 + 273 = 393 K

With this we have the pressure ration of 393/293.

So,F120 = 1.34 APa

6 0

3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

3 years ago

Read 2 more answers