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OLEGan [10]
3 years ago
12

Imagine you’re an engineer making a string of battery powered holiday lights. If a bulb burns out current cannot flow through th

at bulb any linger (like if the wire at the bulb had been cut) . Figure out how to hook up 2 light bulbs and a battery so that when one bulb burns out (or disconnected) the other stays lit.
Physics
1 answer:
Anit [1.1K]3 years ago
4 0

Answer:

The 2 light bulbs can be connected in parallel to each other to avoid disconnection when one bulb burns out.

Explanation:

The parallel connection is required not series. A parallel connection is the connection of electronic components (e.g bulbs, LED, resistors, capacitors etc) in such a way that the same voltage is supplied across the ends of the components.  While in a series connection, the components are connected to each other end-to-end.

As regard the question, parallel connection ensures that the brightness any of the bulbs is not affected with respect to the other bulbs. And other bulbs continue to function when any burns out. The 2 light bulbs should be connected in parallel to the baterry to avoid disconnection of all the bulbs.

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Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0
3 years ago
I need help on 6, 7, 8, and 9
Ivahew [28]
6 is b. part B on 6 is a. 7 is a. partB ON 7 b
5 0
3 years ago
Which object exerts more gravitational force, object A with a mass of 15 grams and a density of 2g/cm3 or object B with a mass o
puteri [66]
The answer is B.

More mass means more gravitational force.

Hope it helps!
8 0
3 years ago
If f(x) = -x2+x-1 then value of f(f(2)) is
qaws [65]

Answer:

<u>-13</u>

Explanation:

<u>Function</u>

  • f(x) = -x² + x - 1

<u>Solving</u>

  • Substitute 2 in place of x
  • f(2) = -(2)² + 2 - 1
  • f(2) = -4 + 2 - 1
  • f(2) = -3
  • f(f(2)) = f(-3)
  • f(-3) = -(-3)² - 3 - 1
  • f(-3) = -9 - 3 - 1
  • f(f(2)) = <u>-13</u>
6 0
1 year ago
What is the wavelength that corresponds to a frequency of 6.00x1014 hz?
kipiarov [429]
The basic relationship between wavelength \lambda, frequency f and speed c of an electromagnetic wave is 
\lambda=  \frac{c}{f}
where c is the speed of light. Substituting numbers, we find:
\lambda=  \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{6.0 \cdot 10^{14} Hz}=5\cdot 10^{-7} m
4 0
3 years ago
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