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3 years ago

How long does it take for the moon to revolve once

Physics

2 answers:

Romashka [77]3 years ago

6 0

It takes the moon to revolve 27 days around the earth
:)

Fiesta28 [93]3 years ago

5 0

It takes the moon about 27 days to revolve on its axis.

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A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down

rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

$V = \sqrt{\frac{T}{\rho}}$

Where,

T = Tension

$\rho =$ Linear density

Our data are given by

Tension , T = 70 N

Linear density , $\rho = 0.7 kg/m$

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

$V = \sqrt{\frac{T}{\rho}}$

$V = \sqrt{\frac{70}{0.7}$

$V = 10m/s$

Speed can also be expressed as

$V = \lambda f$

Re-arrange to find \lambda

$\lambda = \frac{V}{f}$

Where,

f = Frequency,

Which is also described in function of the Period as,

$f = \frac{1}{T}$

$f = \frac{1}{0.35}$

$f = 2.86 Hz$

Therefore replacing to find $\lambda$

$\lambda = \frac{10}{2.86}$

$\lambda = 3.49m$

Therefore the wavelength of the waves created in the string is 3.49m

3 0

3 years ago

How many genders are there?

Dvinal [7]

Answer:2

Explanation:

8 0

3 years ago

Read 2 more answers

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

3 years ago

7.1 Project Guidelines 2021

Ede4ka [16]

I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and

3 0

3 years ago

Consider a monochromatic electromagnetic plane wave propagating in the x direction. At a particular point in space, the magnitud

Scrat [10]

Answer:

a) S = 2.35 10³ J/m²2 ,

b)and the tape recorder must be in the positive Z-axis direction.

the answer is 5

c) the direction of the positive x axis

Explanation:

a) The Poynting vector or intensity of an electromagnetic wave is

S = 1 /μ₀ E x B

if we use that the fields are in phase

B = E / c

we substitute

S = E² /μ₀ c

let's calculate

s = 941 2 / (4π 10⁻⁷ 3 10⁸)

S = 2.35 10³ J/m²2

b) the two fields are perpendicular to each other and in the direction of propagation of the radiation

In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.

the answer is 5

C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis

5 0

3 years ago