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blsea [12.9K]
4 years ago
13

The heating of the filament is what causes the light production (photon emission), and heating is caused by the current in the l

ight bulb. Why isn’t it a good idea to drive light bulbs using constant current sources rather than constant voltage? (Hint: think about the answer to Question 4).
Physics
1 answer:
alexira [117]4 years ago
8 0

Answer:

explained

Explanation:

Yes, the heating of filament is what causes the light production (photon emission), and this heating is caused because of current in the light bulb

(H= i^2*R*t i=current, H= heat, t= time and R= resistance).But using constant current source is not a good idea because in constant current source resistance is very low that can cause short circuit and ultimately fusing it. Whereas in constant voltage source current adjusts itself and prevents fusing because of high resistance in the circuit.

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calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​
igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0
3 years ago
A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c
Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

A = area

I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

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3 years ago
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Answer:

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Explanation:

Couldn't you have used Google???

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Answer:

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