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3 years ago

What are two processes that transfer water into the atmosphere

1 answer:

4 0

Water enters the atmosphere through evaporation, transpiration, excretion and sublimation: Transpiration is the loss of water from plant

Hope this helped you! :D

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If a planet has an eccentricity of 0.92 and a distance of the major axis is 5,000,000 km, then what is the distance between foci

VladimirAG [237]

Answer:

it’s the third option: 4,600,000 km

Explanation:

just took the test and that was correct

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3 years ago

Which statement describes the effects of forces on an object

lukranit [14]

Answer:

Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.

Explanation:

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3 years ago

A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through

aniked [119]

Answer:

$a_2=24.5\ \text{m/s}^2$ towards right

$a_1=2.1875\ \text{m/s}^2$ towards left

Explanation:

$m_1$ = Mass of skater = 56 kg

$m_2$ = Mass of ball = 5 kg

v = Final velocity of ball = 7 m/s

u = Initial velocity of ball = 0

s = Distance the ball moved in the hand of the skater = 1 m

Moving left is considered and moving right is considered positive.

From kinematic equations of motion we have

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2$

So, the ball will move towards right with a magnitude of acceleration $a_2=24.5\ \text{m/s}^2$ .

The force on the ball will be

$F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}$

The force on the ball is $122.5\ \text{N}$

The reaction force on the skater will be equal to the force on the ball but will have opposite direction.

$-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2$

So, the skater will move towards left with a magnitude of acceleration $2.1875\ \text{m/s}^2$

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3 years ago

A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component o

Katen [24]

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, $a=49\ m/s^2$

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

$F=75\ kg\times 49\ m/s^2$

F = 3675 N

Ratio, $R=\dfrac{F}{W}$

$R=\dfrac{3675}{75\times 9.8}=5$

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

$F'=\sqrt{F^2+W^2}$

$F'=\sqrt{(3675)^2+(75\times 9.8)^2}$

F' = 3747.7 N

The direction of force is calculated as :

$\theta=tan^{-1}(\dfrac{W}{F})$

$\theta=tan^{-1}(\dfrac{1}{5})$

$\theta=11.3^{\circ}$

Hence, this is the required solution.

8 0

4 years ago

Fill up the blank :- Weight is measured as the downward force of __________.

In-s [12.5K]

<span>Weight is measured as the downward force of gravity.</span>

3 0

3 years ago