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3 years ago

How does vapor release energy into the atmosphere

Physics

2 answers:

Oduvanchick [21]3 years ago

4 0

I would say a because i just had this

geniusboy [140]3 years ago

3 0

The correct answer is C, Condesation

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For a science project, you would like to horizontally suspend an 8.5 by 11 inch sheet of black paper in a vertical beam of light

liubo4ka [24]

Answer:

I = 3.9 x 10⁷ W/m²

Explanation:

given,

Sheet of black paper dimension = 8.5 x 11 inch

Area of sheet = 8.5 x 11 = 93.5 inch^2

1 inch =0.0254 m

Area = 0.06032 m²

mass of sheet = 0.80 g

Force = m g = 0.8 x 9.8 x 10⁻³ N

= 7.84 x 10⁻³ N

speed of light = c = 3 x 10⁸ m/s

Using equation

$F = \dfrac{IA}{c}$

where I is the intensity of light

$7.84 \times 10^{-3} = \dfrac{I\times 0.06032}{3 \times 10^8}$

$2.352 \times 10^{6} = I\times 0.06032$

$I = \dfrac{2.352 \times 10^{6}}{0.06032}$

I = 3.9 x 10⁷ W/m²

Intensity of the light is equal to I = 3.9 x 10⁷ W/m²

4 0

3 years ago

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

---------------------------------------------------

$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

6 0

3 years ago

Read 2 more answers

1. The center of gravity of both cats and humans is roughly in the thoracic region. However, they each bear their weight very di

Mama L [17]

Answer:

Major, weight-bearing structures are the bones of the body that are strong and dense to be able to bear the weight of the body. The major, weight-bearing structures of cat and human skeletons are :

Human skeleton: The body weight of an individual is on his pelvic girdles that are attached to the bones of lower limbs. Thigh bones, leg bones, and bones of feet comprise lower limbs The lower limbs consist of the thigh, the leg, and the foot.

Cat skeleton: cats are quadrupedal so it bears all the body weight on shoulders and legs that includes the Scapula and pelvis.

6 0

2 years ago

How much does the earth weigh

tatiyna

Answer:

5.972 × 10^24 kg

Explanation:

4 0

3 years ago

Read 2 more answers

Which is an advantage of storing data digitally?