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2 years ago

The astronomer Kepler develop the three laws a planetary motion what is meant by the love distance. Relationship

Physics

1 answer:

Mila [183]2 years ago

4 0

Explanation:

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A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

2 years ago

A farsighted boy has a near point at 2.3 m and requires eyeglasses to correct his vision. Corrective lenses are available in inc

tino4ka555 [31]

Answer:

P = 3.5 D

Explanation:

As we know that convex lens is to be used to make the near point of eye to be correct

So we will have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we have

$d_i = 2.3 m = 230 cm$

$d_o = 25 cm$

now plug in all values into the formula

$-\frac{1}{230} + \frac{1}{25} = \frac{1}{f}$

$f = 28 cm$

now for power of lens

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

$P = \frac{1}{0.28} = 3.5 D$

so the power in dioptre is

P = 3.5 D

5 0

2 years ago

A 6 m/s vector pointing North is added to a 2 m/s vector pointing East. What are the magnitude and direction of the resultant?

irina [24]

Answer:

A + B = C Ax = 2 Ay = 0 Bx = 0 By = 6

Ax + Bx = Cx = 2

Ay + By = Cy = 6

C = (2^2 + 6^2)^1/2 = 6.32

Tan Cy / Cx = 6 / 2 = 3

Cy at 71.6 deg

6 0

3 years ago

Study the diagram. Point C identifies the _____ of the wave.

Sonbull [250]

Answer:

Amplitude : The height of the wave from the origin to the crest/peak or trough

Explanation:

5 0

2 years ago

Read 2 more answers

A ship anchored at sea is rocked by waves that have crests Lim apart the waves travel at 70m/S, at what frequency do the waves r

inessss [21]

Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answer:

0.7 Hz

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave

make f the subject of the equation

f = v/λ................. Equation 2

From the question,

Given: v = 70 m/s, λ = 100 m ( distance between successive crest)

Substitute these values into equation 2

f = 70/100

f = 0.7 Hz

Hence the frequency at which the wave reach the ship is 0.7 Hz

3 0

2 years ago