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HACTEHA [7]
3 years ago
7

Daniela is writing a report on the country of Australia, and she wants to tell how far Australia is from Florida, where she live

s.
Which of the following units of measure would be most appropriate for describing the distance between Florida and Australia?


A. kilometers

B. kilometers/second

C. light-years

D. meters
Physics
1 answer:
o-na [289]3 years ago
8 0

Answer:

A. kilometers

Explanation:

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MATHPHYS CAN U HELP ME PLEASE
ludmilkaskok [199]

Explanation:

(1) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J

The heat added to melt the ice is:

q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J

The heat added to warm the water to 100°C is:

q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J

The heat added to evaporate the water is:

q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J

The heat added to warm the steam to 115°C is:

q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J

The total heat needed is:

q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J

q = 125,654.34 J

(2) When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (6 − 11) + C₂ (25 − 11) = 0

-5 C₁ + 14 C₂ = 0

C₁ = 2.8 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 33) + C₃ (37 − 33) = 0

-8 C₂ + 4 C₃ = 0

C₂ = 0.5 C₃

Substituting:

C₁ = 2.8 (0.5 C₃)

C₁ = 1.4 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(1.4 C₃) (6 − T) + C₃ (37 − T) = 0

(1.4) (6 − T) + 37 − T = 0

8.4 − 1.4T + 37 − T = 0

2.4T = 45.4

T = 18.9°C

(3) Heat gained by the ice = heat lost by the tea

mL + mCΔT = -mCΔT

m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)

m (397372 J/kg) = 1473.472 J

m = 0.004 kg

m = 4 g

4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.

(4) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J

The heat added to melt the ice is:

q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J

The heat added to warm the melted ice to T is:

q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T

The heat removed to cool the water to T is:

q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)

q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T

The heat removed to cool the copper to T is:

q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)

q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T

Therefore:

3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T

13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T

(2266.742 J/°C) T = 44792.58 J

T = 19.8°C

(5) Kinetic energy of the hammer = heat absorbed by ice

KE = q

½ mv² = mL

½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)

m = 9.68×10⁻⁷ kg

m = 9.68×10⁻⁴ g

(6) Heat rate = thermal conductivity × area × temperature difference / thickness

q' = kAΔT / t

q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)

q' = 2943 W

After 10.7 hours, the amount of heat transferred is:

q = (2943 J/s) (10.7 h × 3600 s/h)

q = 1.13×10⁸ J

q = 113 MJ

6 0
3 years ago
A hamster eats a carrot before using its hamster wheel. The hamster wheel is connected to a generator which powers a light bulb.
Andrei [34K]

Answer: Chemical → Mechanical → Electrical → Radiant

Explanation:

First, the Hamster eats the carrot, then the hamster is getting chemical energy.

Now the hamster starts using his wheel, then he "transforms" the chemical energy into mechanical energy.

Now the mechanical energy is connected to a generator, this means that the mechanical energy (the rotation of the wheel) is being converted into electrical energy.

And we know that there is a light bulb powered by this electrical energy, then we have electrical energy being transformed into radiant energy.

Then the correct option is:

Chemical → Mechanical → Electrical → Radiant

6 0
3 years ago
At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 54.0 N at the s
butalik [34]

Answer:

m=5.51Kg

Explanation:

The weight of an object on Earth is given by W=mg, so we can calculate its mass by doing m=W/g, which for our values is:

m=(54N)/(9.8m/s^2)=5.51Kg

<em>Nothing is being asked</em> about Io but if one wanted to know the weight <em>W'</em> of the watermelon there one just have to do:

W'=ma=(5.51Kg)(1.81m/s^2)=9.97N

4 0
3 years ago
A light ray incident on a block of glass makes an incident angle of 50.0° with the normal to the surface. The refracted ray in t
Vladimir [108]

Answer:

The index of refraction of the glass is 1.3

Explanation:

Given data:

i = incident angle = 50°

r = refracted angle = 36.1°

The index of refraction according Snell´s law is:

n=\frac{1*sini}{sinr} =\frac{1*sin50}{sin36.1} =1.3

4 0
3 years ago
9. How much energy does a 100 W light bulb use in half an hour? If the light bulb converts
yulyashka [42]

The energy used by the light bulb in half an hour is 180000 J and the amount of thermal energy generated is 158400 J.

What is Energy?

Energy is the ability or the capacity to do work.

To calculate the energy of the light bulb we use the formula below

Formula:

  • E = Pt.......... Equation 1

Where:

  • E = Energy used by the bulb in a half-hour
  • P = Power of the bulb
  • t = Time

Given:

  • P = 100 W
  • t = 1/2 hour = 30 minutes = (30×60) = 1800 seconds

Substitute these values into equation 1

  • E = (100×1800)
  • E = 180000 J

  • If the light converts 12% of electric energy to light energy, then  88% of the energy is used to generate thermal energy

Therefore,

  • Thermal energy = (180000×88/100) = 158400 J

Hence, the energy used by the light bulb in half an hour is 180000 J and the amount of thermal energy generated is 158400 J.

Learn more about energy here: brainly.com/question/21927255

#SPJ1

6 0
2 years ago
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