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Leokris [45]
3 years ago
10

A candle is an energy converter why??​

Physics
2 answers:
Ludmilka [50]3 years ago
7 0
The heat from the wick melts the wax which gets absorbed in the wick and then gets burnt (which is really oxidation) to produce heat energy as well as light energy. The energy transforms from chemical energy to heat and light energy. Because when the candle burns a chemical reaction occurs and produces heat and light.
Harrizon [31]3 years ago
7 0
I have no clue but I think the other answer is right
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Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?​
earnstyle [38]

Answer:

i hope this will help you :)

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

6 0
4 years ago
During summer vacation, Juan traveled a total distance of 400 miles. His trip took 8 hours.
AVprozaik [17]

Answer:

<em>5</em><em>0</em><em> </em><em>m</em><em>i</em><em>l</em><em>e</em><em>s</em><em>/</em><em>h</em><em>o</em><em>u</em><em>r</em>

Explanation:

his average speed=400/8

=50miles/hour

3 0
3 years ago
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Zigmanuir [339]
The North Star, or Polaris, is the brightest star in the constellation Ursa Minor, the little bear (also known as the Little Dipper). As viewed by observers in the Northern Hemisphere, Polaris occupies a special place
5 0
3 years ago
Based on Dalton’s law of partial pressures if I have 1 L sample of helium at standard temperature has pressure has a volume of 2
sammy [17]

Answer:

wat?.... anyways if 2 + 2 is 4 then what is 10 x 3974793 ?

Explanation:

8 0
2 years ago
Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water,
Nitella [24]

<u>Answer:</u> The amount of heat needed is 29000 Cal.

<u>Explanation:</u>

The process involved in this problem are:

(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)

Now, we calculate the amount of heat released or absorbed in all the processes.

  • <u>For process 1:</u>

q_1=mC_{p,s}\times (T_2-T_1)

where,

q_1 = amount of heat absorbed = ?

m = mass of ice = 200 g

T_2 = final temperature = 0^oC

T_1 = initial temperature = -30^oC

Putting all the values in above equation, we get:

q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal

  • <u>For process 2:</u>

q_2=m\times L_f

where,

q_1 = amount of heat absorbed = ?

m = mass of water or ice = 200 g

L_f = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

q_2=200g\times 80Cal/g=16000Cal

  • <u>For process 3:</u>

q_3=m\times C_{p,l}\times (T_{2}-T_{1})

where,

q_3 = amount of heat absorbed = ?

m = mass of water = 200 g

T_2 = final temperature = 50^oC

T_1 = initial temperature = 0^oC

Putting all the values in above equation, we get:

q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal

Calculating the total heat absorbed, we get:

Q=q_1+q_2+q_3

Q=3000+16000+10000=29000Cal

Hence, the amount of heat needed is 29000 Cal.

7 0
4 years ago
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