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andre [41]
2 years ago
6

the amount of water returning to the earth through precipitation is blank the amount of water leaving the earth through evaporat

ion
Physics
1 answer:
nalin [4]2 years ago
5 0

<u>Answer:</u>

<em>The amount of water entering the earth through precipitation is equal to the amount of water leaving earth through transpiration.</em>

<u>Explanation:</u>

Rates of precipitation and evaporation vary widely according to regions and seasons. But in a global scale the rates are equal. Thus the total amount of earth’s water maintains its constancy even though there is a continuous change in forms of water.

Evaporation and transpiration are the forms in which Water leaves the earth and it returns to the earth in various forms of precipitation like rain, snow, dew, fog etc. This water then reaches ocean and land. The water that reaches the land flows as surface run off into rivers and water bodies or seep into the ground replenishing the ground water table.

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slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa
xenn [34]

Answer:

Maximum height reached by the rocket is

y_{max} = 308 m

total time of the motion of rocket is given as

T = 16.44 s

Explanation:

Initial speed of the rocket is given as

v_i = 50 m/s

acceleration of the rocket is given as

a = 2 m/s^2

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

v_f^2 - v_i^2 = 2 a d

v_f^2 - 50^2 = 2(2)(150)

v_f = 55.68 m/s

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 55.68^2 = 2(-9.81)(y - 150)

y_{max} = 308 m

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

v_f - v_i = at

55.68 - 50 = 2 t

t_1 = 2.84 s

2) time to reach ground from this height

\Delta y = v_y t + \frac{1}{2}gt^2

-150 = 55.68 t - \frac{1}{2}(9.81) t^2

t_2 = 13.6 s

so total time of the motion of rocket is given as

T = 13.6 + 2.84 = 16.44 s

3 0
3 years ago
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
Kipish [7]

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

7 0
3 years ago
you give a shopping cart a job down the aisle the cart is full of groceries and has a mass of 18 kilograms the cart accelerates
mezya [45]

                 Force = (mass) x (acceleration)

                 Force = (18 kg) x (3 m/s²) = 54 newtons

As long as you continue pushing the cart with 54 newtons of force,
it will accelerate at 3 m/s². 

At the instant you release it, or keep your hands on it but stop pushing,
it will stop accelerating.  It'll continue forward at the speed it had when
the 54 newtons of force stopped.
6 0
2 years ago
Read 2 more answers
A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

F=qvB

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

qvB = \frac{mv^2}{r}

which can be rewritten as

v=\frac{qB}{mr}

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

\frac{2\pi r}{T}=\frac{qB}{mr}

So, we get:

T=\frac{2\pi m r^2}{qB}

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) a. f/10

The frequency of revolution of a particle in uniform circular motion is

f=\frac{1}{T}

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}

6 0
3 years ago
A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the hi
Volgvan

Answer: D. 0.29 m

Explanation:

We will use the following equations to describe the leap of the cat:

y=V_{o}sin\theta t-\frac{gt^{2}}{2}   (1)

V_{y}=V_{oy}-gt   (2)

Where:

y  is the height of the cat  

V_{oy}=V_{o}sin\theta is the cat's initial velocity

\theta=60\°

g=9.8m/s^{2}  is the acceleration due gravity

t is the time

V_{y} is the y-component of the velocity

Now the cat will have its maximum height y_{max} when V_{y}=0. So equation (2) is rewritten as:

0=V_{oy}-gt   (3)

Finding t:

t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}   (4)

t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}   (5)

t=0.24 s   (6)

Substituting (6) in (1):

y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}   (7)

Finally:

y_{max}=0.287 m \approx 0.29 m   (8)

3 0
3 years ago
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