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hjlf
3 years ago
15

Which energy transformation occurs when you light a gas burner?​

Physics
1 answer:
Eddi Din [679]3 years ago
3 0

Answer:

Chemical energy to thermal(heat) energy

Explanation:

A gas burner contains chemical hydrocarbon such as methane, this methane is highly flammable.

The gas burner which contains methane when lighted up converts the chemical energy in the chemical to heat energy through the fire it produces. The fire then helps in heating up the surface it is applied to.

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Please answer help me
Andrews [41]
I believe the answer is x
7 0
3 years ago
A rocket takes off from Earth and starts flying to Mars. What happens to the force of gravity between the rocket and Earth as th
lawyer [7]

The force of gravity between Earth and Mars will decrease.

The gravitational law is given as-

           F = G mM/r²

  here,  m= mass of rocket

             M = mass of earth

             r = distance between earth and rocket

So, as rocket takes off from earth and fly towards mars then the distance starts to increase between earth and rocket, and the gravitational pull between them starts to weaken. Then a point will reach when rocket will far from gravity of earth and could probably enter the gravity of Mars.

Learn more about gravitational law here:

 brainly.com/question/12101547

     #SPJ4

5 0
2 years ago
A car accelerates from 20.0 m/s to 28.0 m/s over a distance of 50.0 m. What is the car’s acceleration?
Blababa [14]

Answer:

Explanation:

V^2=V^2_o+2a(x_f-x_i)

28^2=20^2+2a(50)\\784=400+100a\\384=100a\\a=3.84m/s^2

5 0
3 years ago
A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric
sp2606 [1]

Answer:

a)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)E=0

c)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a)    r < a

At a radius r ,from gauss theorem

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)   a < r < b

E.ds=\dfrac{q_i}{\varepsilon _o}

The total induce in this surface = - Q+ Q =0

E.ds=\dfrac{0}{\varepsilon _o}

E = 0

c)   r > b

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

4 0
3 years ago
Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24
Anastasy [175]

Answer:

97 minutes and 12 seconds

Explanation:

Let x be the total time

From the first statement

\frac{10}{100}x=24\times 60+18\\\Rightarrow 0.1x=1458\\\Rightarrow x=\frac{1458}{0.1}\\\Rightarrow x=14580\ seconds

Total time of Dara's exercise is 14580 seconds

40 %

\frac{40}{100}\times 14580=5832\ seconds

\frac{5830}{60}=97.2\ minutes=97+0.2\times 60\\ =97\ minutes\ and\ 12\ seconds=1\ hour\ 37\ minutes\ and\ 12\ seconds

Time when she had completed 40% was 1 hour 37 minutes and 12 seconds

5 0
3 years ago
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