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2 years ago

7

How does water get up to the atmosphere, and how does it get back down to earth surface

1 answer:

IrinaK [193]2 years ago

6 0

Answer:

Water gets up to the Earth's atmosphere by evaporating from a body of water, which is then they become water vapor. It returns back to the surface by returning back to its water state and falling back down (as rain). The water vapor turns into clouds (clouds are really just water droplets), and when it cannot hold anymore waters, it disperses all the water (by raining).

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A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the

laiz [17]

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, $\Delta f = 4\ beats/s$

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

$\Delta f = f - f_{i}$

$f_{i} = f \pm \Delta f$

when

$f_{i} = f + \Delta f = 523 + 4 = 527 Hz$

when

$f_{i} = f - \Delta f = 523 - 4 = 519 Hz$

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

7 0

3 years ago

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The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.55 g of conden

SSSSS [86.1K]

Answer:

m = 62.14 g

Explanation:

Energy used to melt the ice is the energy released by the condensation of the water forms on the glass

so here we have

energy for the condensation of water is given as

let mass of water condensed = m

$E = m_1 L_f$

now the energy of vaporization is given as

$E = m_2 L_v$

here we know that

$L_f = 79.8 kCal/kg$

$L_v = 580 k Cal/kg$

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$8.55 \times 580 = m \times 79.8$

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3 years ago

Now imagine a person dragging a 50 kg box along the ground with a rope, as

ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

= 50 x 9.8

= 490 N

The coefficient of friction,

μ = f/ η

= 30 / 490

= 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0

3 years ago

0.A 20-g bullet moving at 1 000 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s. What is the kinetic

Fiesta28 [93]

Answer:

KE = 0.162 KJ

Explanation:

given,

mass of bullet (m)= 20 g = 0.02 Kg

speed of the bullet (u)= 1000 m/s

mass of block(M) = 1 Kg

velocity of bullet after collision (v)= 100 m/s

kinetic energy = ?

using conservation of momentum

m u = m v + M V

0.02 x 1000 = 0.02 x 100 + 1 x V

20 = 2 + V

V = 18 m/s

now,

Kinetic energy of the block

$KE = \dfrac{1}{2}mv^2$

$KE = \dfrac{1}{2}\times 1 \times 18^2$

KE = 162 J

KE = 0.162 KJ

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3 years ago

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3 years ago

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