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Artist 52 [7]
3 years ago
11

What is the primary energy transformation that occurs when a match is lit? A. Electrical energy is transformed into heat and lig

ht. B. Chemical energy is transformed into heat and light. C. Chemical energy is transformed into electrical energy. D. Heat and light are transformed into chemical energy.
Physics
1 answer:
Lorico [155]3 years ago
6 0

B. chemical energy is transformed into heat and light


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It is important to maintain septic systems in order to <span>prevent untreated sewage from contaminating groundwater</span>
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For states with larger values of n that admit a d subshell, the d subshell is at even higher energy than the p subshell. In fact
frosja888 [35]

Answer:

Explanation:

Correct order of energy of energy states within an atom is as follows

1s , 2 s, 2 p 3 s ,3 p, 4s , 3d,  4p, 4p,5s etc.

There is discrepancies in this order due to overlapping of outer orbitals with inner orbitals so sometimes outer orbitals have lower energy.

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What is the most important safety rule in cycling?
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An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0
Murrr4er [49]

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

8 0
2 years ago
Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge
NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is V=\frac{1}{4\pi\epsilon} \frac{q}{d}

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0
3 years ago
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