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Licemer1 [7]
2 years ago
14

11. Which type of cloud is shown in this image?

Physics
1 answer:
torisob [31]2 years ago
5 0

Cumulus, cirrus, and stratus clouds are the three types of clouds shown in this photograph.

<h3>How does a cloud behave in the sky?</h3>

The importance of clouds cannot be overstated. One of the causes is precipitation, such as rain or snow.

Clouds make the surface warmer at night by reflecting heat back to it. Clouds may shield us from the sun and make Earth cooler throughout the day.

The type of cloud is shown in this image is;

1.Cumulous

2.Cirrus

3.Stratus

1. Cumulous

The fluffy, white, cotton-top clouds known as cumulus have such a gentle appearance that you may imagine angels relaxing and going about their business on them.

2. Cirrus

High clouds in the cirrus genus are comprised of ice crystals. The characteristic appearance of cirrus clouds is fragile and wispy with white filaments.

3. Stratus

Contrary to convective or cumuliform clouds, which are produced by rising thermals, stratus clouds are low-level clouds with horizontal stratification and a homogenous base.

Hence, the type of clouds shown in this image is cumulus, cirrus, and stratus.

To learn more about the cloud refer;

brainly.com/question/18370044

#SPJ1

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Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co
mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copperV=8.0\times10^{-6}\ m^3

Volume of liquid V=7.6\times10^{-6}\ m^3

Latent heat of fusion L= 13.05 kJ

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}

Put the value into the formula

\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}

\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}

\Delta T=4.162\ K

Hence, The change in the equilibrium melting point is 4.162 K.

5 0
3 years ago
Block 1, of mass m1 = 2.70 kg , moves along a frictionless air track with speed v1 = 27.0 m/s . It collides with block 2, of mas
ki77a [65]

Answer:

a) Block 1 = 72.9kgm/s

Block 2 = 0kgm/s

b) vf = 1.31m/s

c) ∆KE = 936.36Joules

Explanation:

a) Momentum = mass× velocity

For block 1:

Momentum = 2.7×27

= 72.9kgm/s

For block 2:

Momentum = 53(0) (body is initially at rest)

= 0kgm/s

b) Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses of the block

u1 and u2 are their initial velocity

v is the common velocity

Given m1 = 2.7kg, u1 = 27m/s, m2 = 53kg, u2 = 0m/s (body at rest)

2.7(27)+53(0) = (2.7+53)v

72.9 = 55.7v

V = 72.9/55.7

Vf = 1.31m/s

c) kinetic energy = 1/2mv²

Kinetic energy of block 1 = 1/2×2.7(27)²

= 984.15Joules

Kinetic energy of block 2 before collision = 0kgm/s

Total KE before collision = 984.15Joules

Kinetic energy after collision = 1/2(2.7+53)1.31²

= 1/2×55.7×1.31²

= 47.79Joules

∆KE = 984.15-47.79

∆KE = 936.36Joules

7 0
3 years ago
Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
4 years ago
Consider being at the top of cliff and throwing a book off the ledge. The book leaves at an angle of 52 degrees and a velocity o
Masja [62]

The vertical distance through which the book falls is determined as 1,048.8 m.

<h3>Height of the book fall</h3>

The vertical distance through which the book falls is calculated as follows;

h = vt + ¹/₂gt²

where;

  • h is height of fall
  • v is initial vertical velocity
  • g is acceleration due to gravity

h = (16 x sin52)(13.4)  + (0.5)(9.8)(13.4²)

h = 1,048.8 m

Thus, the vertical distance through which the book falls is determined as 1,048.8 m.

Learn more about height of fall here:  brainly.com/question/15611384

#SPJ1

3 0
2 years ago
Substances in a mixture keep their own properties.<br> True False
ra1l [238]
That is true. They aren't undergoing a chemical change. They are still the same.
3 0
3 years ago
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