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hoa [83]
3 years ago
7

What are 5 uses of reflection?

Physics
2 answers:
dangina [55]3 years ago
8 0
Optomety-contact/corrective lenses
binoculars periscopes and refracting telescopes and refracting telescopes
mangnifying glass
knowledge of refraction creates the most sparkly diamonds
prisms to divdide white light into it's into  it's component colours
certain optical illusions

Gre4nikov [31]3 years ago
3 0
<span>here u go!
fiber-Optics
Mirrors
Sonar
Radar
Study of seismic waves </span>
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Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second. It hits stationary ball B and they undergo
blagie [28]
<span>If Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second, and It hits stationary ball B and they undergo elastic collision, thus the two balls have different masses, then the following statement which is true is the statement that stated that there was no y-momentum initially.</span>
7 0
3 years ago
The potential energy of a catapult was completely converted into kinetic energy by releasing a small stone with a mass of 20 gra
DerKrebs [107]

1,000 grams = 1 kilogram
20 grams = 0.02 kilogram

Kinetic energy = (1/2) (mass) x (speed)²

                           (1/2) (0.02) x (15)² =

                                 (0.01)  x  (225)  =  2.25 joules
 
4 0
3 years ago
Read 2 more answers
Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0
3 years ago
Identify five common business uses for electronic spreadsheets
Sloan [31]


For the answer to the question above
Forecasting how a business might do in the future. 
Calculating tax. 
Doing basic payrolls. 
Calculating Revenues. 
Producing charts. 

--Going past 5-- 

Inventory tracking 
Very (VERY) basic CRM for small businesses 
I hope my answer helped you. 
4 0
3 years ago
Read 2 more answers
A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur
emmasim [6.3K]

• Net vertical force on the block:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

(<em>n</em> = magnitude of normal force, <em>w</em> = weight)

<em>n</em> = <em>w</em> = <em>m g</em>

(<em>m</em> = mass, <em>g</em> = 9.8 m/s²)

<em>n</em> = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ <em>F</em> = -<em>f</em> = <em>m a</em>

(<em>f</em> = mag. of friction, <em>a</em> = acceleration)

We have <em>f</em> = <em>µ</em> <em>n</em> = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) <em>a</em>

<em>a</em> = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆<em>x</em> in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆<em>x</em>

∆<em>x</em> = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

8 0
3 years ago
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