All categories

2 years ago

What pattern is seen on the moon from earth?

Physics

1 answer:

igomit [66]2 years ago

7 0

If you mean pattern designs it’s rocky bumpy and you can see the circle patterns

If you mean phases the here’s a picture

You might be interested in

ln the constellation of Orion, you can see a group of stars and other objects that appear in the shape of a sword. Ln the middle

bagirrra123 [75]

Answer:

Nebula

Explanation:

Given that in the constellation of Orion, you can see a group of stars and other objects that appear in the shape of a sword. Ln the middle of the sword, a bright "fuzzy star" appears. Astronomers looking at this object through telescopes refer to it as a "stellar nursery." Another name for this object is called NEBULA.

The space where new stars are forming anew is known as nebulae

7 0

3 years ago

Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart

Romashka [77]

Answer:

E = 5291.00 N/C

Explanation:

Expression for capacitance is

$C = \frac{\epsilon A}{d}$

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

$C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}$

$C = 24.06 \epsilon$

$C = 24.06\times 8.854\times 10^{-12} F$

$C =2.1\times 10^{-10} F$

We know that capacitrnce and charge is related as

$V = \frac{Q}{C}$

$= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}$

v = 9.523 V

Electric field is given as

$E = \frac{V}{d}$

= $\frac{9.52}{1.8*10^{-3}}$

E = 5291.00 V/m

E = 5291.00 N/C

5 0

2 years ago

The electric field 30cm from a van de Graaff generator is measured to be 28,300N/C. What is the charge of the van de Graaf?

gizmo_the_mogwai [7]

Answer:

Explanation:

EWAN KO LANG DIN BASTA YAN ALAM KO

8 0

2 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago

Convert 93.6 miles per hour. Convert this to kilometers per hour.

Readme [11.4K]

Answer:

150.6 km

Explanation:

One mile is about 1.61 km so multiply 93.6 by 1.6 which gives you above 150.6

3 0

3 years ago