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3 years ago

Can anyone help me on this problem, I’m confused. Thanks

Physics

1 answer:

gregori [183]3 years ago

4 0

The answer would be C. velocity. When white light enters a prism it is refracted. This refraction is caused due to a change in velocity (generally speaking, velocity is considered speed in some cases) when entering the prism. In other words, when light enters a prism at an angle, the glass slows down the light causing the light to split into different colors.

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A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co

Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

m₁ = 0.973kg

when 1.0kg of aluminium block is used, the final temperature of the mixture will be T

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be

heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T = 418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T = 104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0

3 years ago

The simplest form of a substance

likoan [24]

Answer :

C) Atom

Actual answer should be element but depends on the way you interpret the question and the options given to answer it

Explanation :

Atom is the most basic unit of any substance and is what molecules are made of.

Hope it helps if it does let me know by thanking

5 0

3 years ago

In avarage,How many times do a child breathe in a

ollegr [7]

On an approximate scale, A child breaths 20 times a minute as compared to only 12 to 16 in resting phase of an Adult.

So, In 60 minutes (1 hour), They breathe = 20 * 60 = 1200
In 24 hours (1 day), They breathe = 1200 * 24 = 28,800

In short, Your Answer would be: 28,800

Hope this helps!

5 0

2 years ago

Read 2 more answers

a runner starts from rest and has an acceleration of 3 m/s^2. How fast is she running after 1.1 seconds

grigory [225]

Acceleration x time = velocity

Since you're given acceleration and time, just plug the values into the equation.

3 $\frac{m}{s^{2}}$ x 1.1 s = ?

Solve that equation, and remember your velocity should be in m/s.

8 0

2 years ago

A cubical block of iron 10 cm on each side is floating on mercury in a vessel. (i) What is the height of the block above the mer

Gre4nikov [31]

Answer:

i 5.3 cm ii. 72 cm

Explanation:

We know upthrust on iron = weight of mercury displaced

To balance, the weight of iron = weight of mercury displaced . So

ρ₁V₁g = ρ₂V₂g

ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?

V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³

So, the height of iron above the mercury is h = V₂/area of base iron block

= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm

ρ₁V₁g = ρ₂V₂g

ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?

V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ = 7200 cm³

So, the height of column of water is h = V₃/area of base iron block

= 7200 cm³/10² cm² = 72 cm

7 0

3 years ago