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erastova [34]
3 years ago
12

Suppose you saw a waxing gibbous moon. What phase would you expect one week later?

Physics
1 answer:
sattari [20]3 years ago
3 0

The gibbous moon spends roughly a week waxing, from the First Quarter until the Full Moon.  (This is the second week following a New Moon.)

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The surface of the Earth changes from processe such as erosion. Which of these changes to Earth's surface is an example of erosi
Ad libitum [116K]

Answer:

D. the wind picking up dust and carrying it

Explanation:

Erosion is a process in which an agent transfer the top soil to another region, thereby exposing the lower soil. These agents have the ability to move the top layer of soil and deposit it at another place. The major agents in this case are; a running or flowing body of water and wind.

Therefore, the change to the Earth's surface that is an example of erosion is the wind picking up dust and carrying it. Thereby exposing the lower layers.

6 0
3 years ago
Difference between electrolytes and nonelectrolytes.
Trava [24]

Answer:

Electrolytes are salts or molecules that ionize completely in solution. As a result, electrolyte solutions readily conduct electricity. Nonelectrolytes do not dissociate into ions in solution; nonelectrolyte solutions do not, therefore, conduct electricity

Explanation:

8 0
2 years ago
Read 2 more answers
2. Is this chemical equation balanced?<br> 2C4H10 (g) + 13O2 (g) = 8CO2 (g) → 10H2O (l)
ehidna [41]
Yes that is a balaned equation
3 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
Darya [45]

The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

\implies h=v\sqrt{\dfrac hg}

\implies v=\sqrt{hg}

8 0
3 years ago
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