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3 years ago

10

Wich sphere of the earth system includes snowfields ice caps and frozen ground

1 answer:

prisoha [69]3 years ago

5 0

The answer of the given question above would be CRYOSPHERE. The sphere of the Earth system that includes snowfields, ice caps, and frozen ground is called the cryosphere. This is the sphere of the Earth which is the frozen part and are found in the areas of<span> waters surrounding Antarctica and the Arctic. Hope this helps.</span>

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What is the height of the cliff?

Lerok [7]

To find the height of the cliff
use the kinematical equation
height = 1/2*a*t^2
= 5*0.75*0.75 = 2.8125 m

3 0

3 years ago

I don’t not get anything to do with this question

Scorpion4ik [409]

Answer:

C

Explanation:

Mechanical energy = kintetic energy + grav potential energy

ME = KE + GPE <==== re-arrange

GPE = ME - KE <==== KE = 1/2 m v^2

<u>GPE = ME - 1/2 mv^2 </u>

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1 year ago

Read 2 more answers

A proton moves at constant velocity in the +y direction, through a region in which there is an electric field and a magnetic fie

Ivenika [448]

Answer:

a) F_{e} - F_{m} = 0, b) v = 666.67 m / s

Explanation:

For the proton to move y-axis the sum of the electric and magnetic force must be zero, therefore

$F_{e} - F_{m}$ = 0

a) ∑ F = 0

F_{e} - F_{m} = 0

b) we write the forces

q E = q v B

v = E / B

Let's calculate

v = 300 / 0.45

v = 666.67 m / s

4 0

2 years ago

What is the net work done on the 20kg block while it moves the 4 meters?

Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

Weight = 20 x 9.81

Weight = 196.2 N

2.- Calculate the work done

Work = 196.2 x 4

Work = 784.8 J

5 0

3 years ago

The California sea lion is capable of making extremely fast, tight turns while swimming underwater. In one study, scientists obs

anygoal [31]

Answer:

Acceleration of Sea Lion is 4.41 g

This is 49% of maximum jet acceleration given as a = 9g

Explanation:

As we know that the radius of the circular loop is given as

R = 0.37 m

The speed of the fish is given as

$v = 4 m/s$

Now the centripetal acceleration of the sea lion is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{4^2}{0.37}$

$a_c = 43.2 m/s^2$

as we know that

$g = 9.8 m/s^2$

so we have

$a = 4.41 g$

Now Percentage of this acceleration wrt maximum jet acceleration is given as

$P = \frac{4.41 g}{9g} \times 100$

$P = 49$ %

6 0

3 years ago