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3241004551 [841]
4 years ago
14

How is an image produced by a plane mirror different than an image produced by a convex mirror

Physics
2 answers:
KengaRu [80]4 years ago
6 0
The light that is reflected always from you conCAVE or conVEX meaning the light that hits the mirror reflects away from you while a reflection of a plane mirror is directly reflected image without deviation
puteri [66]4 years ago
6 0

Answer:

When light impacts a plane mirror, the light ray is reflected with the same angle as the incident light ray.

For the curved mirrors, like the convex mirror, the normal surface of the mirror is not plane, so the angles change, for example, a light ray that includes horizontally in a curved part (not in the middle, a little bit above it for example) will impact a curved surface, and the reflected light ray will have a direction that "goes away" from the center of the mirror.

This means that those mirrors produce an image like a plane mirror right in the middle, and as you go to the sides, the image starts to be dilated.

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Winds transfer energy in the form heat from the air to the ground!
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Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

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Fronts are termed by the temperature of the air mass that overtakes another air mass. A. True B. False
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16 grams of ice at –32°C is to be changed to steam at 182°C. The entire process requires _____ cal. Round your answer to the nea
ICE Princess25 [194]

Answer:

12432 cal.

Explanation:

The process to change ice at -32 ºC to steam at 182 ºC can be divided into 5 steps:

1. Heat the ice to 0 ºC, which is the fusion temperature.

2. Melt the ice (obtaining liquid water), which is a process at constant pressure and temperature, so the liquid obtained is also at 0ºC.

3. Heat the liquid water from 0 ºC to 100 ºC, which is the vaporization normal temperature of the water.

4. Vaporization of all the water; this is also a process that occurs at constant pressure and temperature, so the produced steam will be at 100ºC.

5. Heat the steam from 100 ºC to 182 ºC.

Each process has a required energy, and the sum of the energy required for each and all of the steps is the total amount of energy required for the whole process:

E_T=E_1+E_2+E_3+E_4+E_5

E_1 is a heating process for the ice, so we know that the energy required is proportional to the temperature difference through the specific heat:

E_1=m*Cp_{sol}*(T_2-T_1)\\E_1=16g*0.5\frac{cal}{gC}*(0-(-32))=256cal

E_2 is a phase change process, so we do not use the specific heat (sensible heat), but the fusion heat (latent heat), so:

E_{2}=m*dh_{f}={16g*80\frac{cal}{g}}=1280cal

Analogously,

E_3=m*Cp_{liq}*(T_3-T_2)=16g*1.00\frac{cal}{gC}*(100-0)C = 1600 cal

E_{4}=m*{dh_{vap}}\\\\E_4=16g*540\frac{cal}{g} =8640cal

E_{5}=m*Cp_{vap}*(T_{5}-T_{4})\\E_{5}={16g*0.5\frac{cal}{gK}*(182-100)K}=656cal

Finally, the total energy required is:

E_T=256cal+1280cal+1600cal+8640cal+656cal\\E_T=12432cal

8 0
3 years ago
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