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3 years ago

List two ways that waves erode land

2 answers:

8 0

Waves erode through forcing air bubbles into tiny cracks in rocks (hydraulic action), and also by abrasion, whereby it causes rocks to be hurled against rock faces.

I hope this helps!

Veseljchak [2.6K]3 years ago

8 0

Answer:

Waves erode by abrasion and deposition

Explanation:

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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

The velocity of an object includes its speed and what

SCORPION-xisa [38]

I believe that the correct answer you are looking for is the distance traveled

5 0

3 years ago

Read 2 more answers

Which type of telescope does not require darkness in order to be able to use it?

exis [7]

Answer:

A type of telescope that does not require darkness in order to be able to use it is the refracting telescope

Explanation:

A refracting telescope consists of a lens and an eyepiece collects light which is then focused to present a magnified, bright and clear image.

The incident light on a refracting telescope is bent by refraction such that the light is focused to the focal point.

In refracting telescopes, the image is formed by bending light, that is by refraction.

The refracting telescope technology has been applied to binoculars and camera zoom lenses.

5 0

3 years ago

At a given instant, the force on an electron is in the +z-direction (out of the page), which the electron is moving in the +x-di

Hitman42 [59]

Answer:

The direction of the B-field is in the +y-direction.

Explanation:

The corresponding formula is

$F_B = qv\times B$

This means, we should use right-hand rule.

Our index finger is pointed towards +x-direction (direction of velocity),

our middle finger should point towards the direction of the B-field,

and our thumb should point towards the +z-direction (direction of the force).

Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.

$\^{x} \times \^{y} = \^{z}$

3 0

3 years ago

Assertion – Reason

Norma-Jean [14]

u have not displayed the question fella but i can guess the answer

Explanation:

the answer is (i)

because the current in a circuit is always directly proportional to its voltage thus making it to make a straight line in its graph

3 0

2 years ago