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3 years ago

A form of erosion in which particles of sand or dust rub across the surface of rocks.

Physics

2 answers:

gayaneshka [121]3 years ago

7 0

The answer is B) Abrasion.

Likurg_2 [28]3 years ago

5 0

Your answer will be Abrasion

Hope i helped have a wonderful day.

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A ball is thrown horizontally from the top of a 41 m vertical cliff and lands 112 m from the base of the cliff. How fast is the

yanalaym [24]

Answer:

4.78 second

Explanation:

given data

vertical cliff = 41 m

height = 112 m

solution

we know here time taken to fall vertically from the cliff = time taken to move horizontally ..........................1

so we use here vertical component of ball

and that is accelerated motion with initial velocity = 0

so we can solve for it as

height = 0.5 × g × t² ........................2

put here value

112 = 0.5 × 9.8 × t²

solve it we get

t² = 22.857

t = 4.78 second

ball thrown horizontally from the top of the cliff in 4.78 second

4 0

3 years ago

He discovered that the orbits of planets are ellipses.

bezimeni [28]

Answer:

Kepler

Explanation:

Kepler discovered that the orbits of planets are ellipses.

6 0

3 years ago

Read 2 more answers

What is the fundamental unit for measuring mass in the metric system ?

goldenfox [79]

Kilogram(kg)
It's not the SI unit of mass in the metric system however.

6 0

2 years ago

Read 2 more answers

Absolute Zero is when energy / molecules stop moving.

denis-greek [22]

Answer:

yes

Explanation:

When all of the molecules (or atoms) in a system stop moving completely, that's as cold as they can get

3 0

2 years ago

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$

Therefore, the separation between the two charges (r) = 1.51 km

3 0

2 years ago