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Leokris [45]
3 years ago
15

A light bulb converts electrical energy into light energy and also some thermal energy. The amount of electrical energy is 50 Jo

ules before the conversions.
Physics
1 answer:
Kitty [74]3 years ago
8 0
If the amount of electrical energy is 50 Joules before the conversions, then it would be 50 Joules after the conversion.

According to law of conservation of energy, we cannot create or destroy energy so it remains constant

Hope this helps!
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At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar
swat32

Answer:

C=1,25\cdot 10^{5} kJ/^{\circ}C

Explanation:

First of all let's define the specific molar heat capacity.

C = \frac{-Q}{n\cdot \Delta T} (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system  

Now, we can find n with the molar mass (M) the mass of the compound (m).

n=\frac{m}{M}=6.95\cdot 10^{-3} moles      

Using (1) we have:

C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}

C=1,25\cdot 10^{5} kJ/^{\circ}C

I hope it helps!

6 0
3 years ago
2) A car travels 5 miles north, and then 10 miles south. What is the person's DISTANCE and
Liono4ka [1.6K]

Answer:

distance traveled is 15 mi

displacement is 5 mi

Explanation:

Distance takes time into account and adds up all the tiny displacements during the entire period of the trip.

Displacement ignores time and looks only at the change in position from the starting point to the ending point.

6 0
3 years ago
Select the statement that best describes how the forces relate. (2 points)
Anarel [89]
"D. Magnetic and electrical forces are similar because they are both related to the interactions between charged particles" best describes how the forces relate.
4 0
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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

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Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

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3 years ago
The most common liquid on planet earth is
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The most common liquid on planet earth is water
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