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qaws [65]
3 years ago
6

When a light wave enters into a medium of different optical density,

Physics
1 answer:
olga nikolaevna [1]3 years ago
6 0
I think it should be option (b)
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1. Place these unknown pH test papers in order from most acidic to most alkaline.
Naddika [18.5K]

Answer: B

Explanation: look at the chart, easy

7 0
2 years ago
If chilled coke and hot tea are <br> kept together tea cools down but ko gets warm why<br>​
olga55 [171]
Tea gets cool beacuse of heat tranfer through convenction
Heat transfers from a area of hot region to cold.
The tea is hot in this case, due to taht factor the heat will move towards teh cooler region which is coke and will make the coke warm


Hope this helped!
5 0
3 years ago
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0
3 years ago
A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W
labwork [276]

Answer:

<h2>650W/m²</h2>

Explanation:

Intensity of the sunlight is expressed as I  = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

10cm = (10/100)m

10cm = 0.1m

100cm² = 0.1m * 0.1m = 0.01m²

Area (in m²) = 0.01m²

Required

Intensity of the sunlight I

I = P/A

I = 6.5/0.01

I = 650W/m²

Hence, the intensity of the sunlight in W/m² is 650W/m²

4 0
3 years ago
A swimming pool is an example of an open system. The pool loses 10,500 J
zaharov [31]

Answer: A a decrease of 8000

Explanation: 10,500 - 2,500= 8000

8 0
3 years ago
Read 2 more answers
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