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3 years ago

6

I dnt know how to do it

1 answer:

NNADVOKAT [17]3 years ago

4 0

Here's what you need to know about waves:

Wavelength = (speed) / (frequency)

Now ... The question gives you the speed and the frequency,
but they're stated in unusual ways, with complicated numbers.

Frequency: How many each second ?
The thing that's making the waves is vibrating 47 times in 26.9 seconds.
Frequency = (47) / (46.9 s) = 1.747... per second. (1.747... Hz)

Speed: How far a point on a wave travels in 1 second.
The crest of one wave travels 4.16 meters in 13.7 seconds.
Speed = (4.16 m / 13.7 sec) = 0.304... m/s

Wavelength = (speed) / (frequency)

Wavelength = (0.304 m/s) / (1.747 Hz) = 0.174 meter per second

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State in which atoms or molecules are very close together and are regularly arranged

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Solid, because the atoms maintain their form and do not take the shape of their container. The particles share tight, close bonds due to this set shape

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3 years ago

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A circular coil of wire, with N turns of radius a, is located in the field of an electromagnet. The magnetic field is perpendicu

weeeeeb [17]

Answer:

they are going to attract

Explanation:

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3 years ago

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

Two balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L

olga_2 [115]

Answer:

Explanation:

Given

mass of balls $m= 5 kg$

$N=45.6 rev/s$

angular velocity $\omega =2\pi N=286.55 rad/s$

Length of Rod $2L=1.1 m$

Tension in the Second half of rod

$T_2=m\omega ^2(2L)=2m\omega ^2L$

$T_2=5\times (286.55)^2\times 1.1$

$T_2=451.609 kN$

For First Part

$T_1-T_2=m\omega ^2L$

$T_1=T_2+m\omega ^2L$

$T_1=3 m\omega ^2L$

$T_1=3\times 5\times (286.55)^2\times 0.55$

$T_1=677.41 kN$

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3 years ago

Who were Socrates and Plato

VARVARA [1.3K]

Socrates was a classical Greek philosopher and Plato was a philosopher in Classical Greece

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4 years ago

Read 2 more answers