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4 years ago

How does water change the shape of earth’s surface?

Physics

2 answers:

Dominik [7]4 years ago

8 0

Answer:

Water shapes Earth’s surface by wearing away soil and rock and through the movement of glaciers. Weathering through the action of water and frost or ice also changes the shape of Earth’s surface

Explanation: Sample answer

iren [92.7K]4 years ago

3 0

Water moving across the earth in streams and rivers pushes along soil and breaks down pieces of rock in a process called erosion. ... Water moving in ocean waves carries sand, shells and debris away from some coastal areas and deposits them in new areas, changing the shape of the coastline.

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Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then

Andreas93 [3]

Answer:

The displacement is $\Delta H = - 1 \ m$

The distance is $D = 4 \ m$

Explanation:

From the question we are told that

The height from which the ball is dropped is $h = 1 \ m$

The height attained at the first bounce is $h_1 = 0.8 \ m$

The height attained at the second bounce is $h_2 = 0.5 \ m$

The height attained at the third bounce is $h_3 = 0.2 \ m$

Note : When calculating displacement we consider the direction of motion

Generally given that upward is positive the total displacement of the ball is mathematically represented as

$\Delta H = (0 - h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)$

Here the 0 show that there was no bounce back to the point where Billy released the ball

$\Delta H = (0 - 1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)$

=> $\Delta H = - 1 \ m$

Generally the distance covered by the ball is mathematically represented as

$D = h + 2h_2 + 2h_3 + 2h_3$

The 2 shows that the ball traveled the height two times

$D = 1 + 2* 0.8 + 2* 0.5 + 2* 0.2$

=> $D = 4 \ m$

5 0

3 years ago

52. How did people spend Life in the<br>hunting age-

kondor19780726 [428]

Answer:

what is the hunting age? i've never heard of it

7 0

4 years ago

Read 2 more answers

A particle with charge q and momentum p, initially moving along the x-axis, enters a region where a uniform magnetic field* B=(B

VMariaS [17]

Answer:

Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

So, from the question, we are given that the charge = q, the momentum = p.

=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."

Momentum,P = mass × Velocity, v -----(1).

We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.

Radius,r = P × v / B0 -----------------------------(2).

Centrepetal force = q × B0 × v. ----------(3).

(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.

Equating Equation (2) and (3) gives;

P = B0 × q × r.

Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

4 0

3 years ago

When you rub a balloon on a sweater or your hair, the what is transferred to the balloon giving it a what charge. The balloon wi

Elis [28]

Answer:

When you rub a balloon on a sweater or your hair, the <u>electrons</u> are transferred to the balloon giving it a <u>negative</u> charge. The balloon will then <u>repel</u> any object that has a different charge.

hope this helps :)

5 0

3 years ago

You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in

jekas [21]

$\begin{array}{ccc}&\text{Radius} & \text{Drift Speed}\\d) & 2 \; r &2.5 \; v\\a) & 3 \; r &1 \; v\\b) & 4 \; r &0.5 \; v\\c) & 1 \; r &5 \; v\\\end{array}$