All categories

3 years ago

The diagram below shows the movement of matter in a portion of the water cycle.

Physics

1 answer:

Alisiya [41]3 years ago

7 0

Answer:

d) precipitation

Hope it helps you

And if you want to, pls mark it as the brainliest answer

You might be interested in

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo

denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

$\mu _smg=610$

$\mu _s\times 93\times 9.8=610$

$\mu _s=0.669$

5 0

3 years ago

Please help me !! i’ll mark brainliest if you’re right!

navik [9.2K]

Option 2nd is the answer

3 0

2 years ago

Which statements did both Aristotle and Ptolemy assume? Check all that apply.

tensa zangetsu [6.8K]

Bodies in space traveled in circles.

The planets revolved around the Earth.

6 0

3 years ago

Read 2 more answers

What is the magnitude of a vector with components (15 m, 8 m)?

il63 [147K]

The answer is 17 m because you have to add the 15 m and the 8 m together to get the answer so it will be like this 17x17 = 15x15 + 8x8 got it?

8 0

3 years ago

Read 2 more answers

A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i

cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

$\boxed{\mathfrak{range = \frac{ u {}^{2} \sin 2\theta }{g} }}$

u is the velocity during its take off and $\theta$ is the angle at which its thrown

Given that

u = 8m/ s
$\theta$ = 40°

calculating range using the above formula

$= \frac{ {8}^{2} \sin2(40) }{10}$

$= \frac{64 \times \sin(80) }{10}$

value of sin 80 = 0. 985

$= \frac{64 \times 0.985}{10}$

$= \frac{63.027}{10}$

$= 6.3027$

Hence,

$\mathfrak { \blue{the \: teammate \: is \: \red{\underline{6.3027 \: meters} }\: away } }$

7 0

3 years ago