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Ludmilka [50]
3 years ago
11

Two sets of Christmas lights are available. For set A, when one bulb is removed, the remaining bulbs remain illuminated. For set

B, when one bulb is removed, the remaining bulbs do not operate. Explain the difference in wiring for the two sets.??(a) Two resistors are connected in series across a battery. Is the power delivered to each resistor (i) the same or (ii) not necessarily the same? (b) Two resistors are connected in parallel across a batter. Is the power delivered to each resistor (i) the same or (ii) not necessarily the same.??
Physics
1 answer:
Paul [167]3 years ago
7 0

Explanation:

Bulbs are nothing but resistors that glow when current passes through them.

In Set A, the bulbs (resistors) are connected  parallely to each other, this means that even if one of the bulbs fuses or removed, the circuit will still be completed and others continue to glow.

And in parallel connection if the resistance of the two resistors are same  powered delivered to each is same.

In Set B, bulbs are in series connection, this means that when one of the bulb is removed or fuses, the circuit will break and other bulbs can not operate.In this situation as well  if the resistance of two resistors is same then the power delivered is same.

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poizon [28]

Answer:

2

Explanation:

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2 years ago
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Under the assumption that the beam is a rectangular cantilever beam that is free to vibrate, the theoretical first natural frequ
BartSMP [9]

Answer:

a) Δf = 0.7 n , e)   f = (15.1 ± 0.7) 10³ Hz

Explanation:

This is an error about the uncertainty or error in the calculated quantities.

Let's work all the magnitudes is the SI system

The frequency of oscillation is

        f = n / 2π L² √( E /ρ)

where n is an integer

Let's calculate the magnitude of the oscillation

       f = n / 2π (0.2335)² √ (210 10⁹/7800)

       f = n /0.34257 √ (26.923 10⁶)

       f = n /0.34257    5.1887 10³

       f = 15.1464 10³ n

a) We are asked for the uncertainty of the frequency (Df)

       Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ

in this case no  error is indicated in Young's modulus and density, so we will consider them exact

       ΔE = Δρ = 0

       Δf = df /dL  ΔL

       df = n / 2π   √E /ρ   | -2 / L³ | ΔL

       df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³

       df = n 0.649

Absolute deviations must be given with a single significant figure

        Δf = 0.7 n

b, c) The uncertainty with the width and thickness of the canteliver is associated with the density

 

In your expression there is no specific dependency so the uncertainty should be zero

The exact equation for the natural nodes is

          f = n / 2π L² √ (E e /ρA)

where A is the area of ​​the cantilever and its thickness,

In this case, they must perform the derivatives, calculate and approximate a significant figure

        Δf = | df / dL | ΔL + df /de  Δe + df /dA  ΔA

        Δf = 0.7 n + n 2π L² √(E/ρ A) | ½  1/√e | Δe

               + n / 2π L² √(Ee /ρ) | 3/2 1√A23  |

the area is

        A = b h

        A = 24.9  3.3  10⁻⁶

        A = 82.17 10⁻⁶ m²

        DA = dA /db ΔB + dA /dh Δh

        dA = h Δb + b Δh

        dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³

        dA = (3.3 + 24.9) 0.005 10⁻⁶

        dA = 1.4 10⁻⁷ m²

let's calculate each term

         A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe

         A ’= n/ 2π L² √ (E /ρ)      | ½ 1 / (√e/√ A) |Δe

        A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³

        A '= 0.0266  n

        A ’= 2.66 10⁻² n

       A ’’ = n / 2π L² √ (E e /ρ) | 3/2  1 /√A³ |

       A ’’ = n / 2π L² √(E /ρ) √ e | 3/2  1 /√ A³ | ΔA

       A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷

       A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴

       A ’’ = 6,738 10²

we write the equation of uncertainty

     Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)

The uncertainty due to thickness is

    Δf = 3 10⁻² n

The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small

     Δf = 7 10² n

 d)    Δf = 7 10² n

e) the natural frequency n = 1

       f = (15.1 ± 0.7) 10³ Hz

7 0
3 years ago
The Gaia hypothesis is an example of _____
Fofino [41]
A complex entity involving the Earth's biosphere, atmosphere, oceans, and soil; the totality constituting a feedback or cybernetic system which seeks an optimal physical and chemical environment for life on this planet
4 0
3 years ago
A scientist carries out an experiment. How could she help other scientists judge the validity of her results?
NNADVOKAT [17]
One way to do it is she could right down the data that she got
4 0
3 years ago
Read 2 more answers
A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___
34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8·\hat j, in vector form

2) 1.0 km East = 1.0·\hat i, in vector form

3) 1.6 km South = -1.6·\hat j, in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

Final displacement, = 1.0·\hat i + 1.2·\hat j

Where;

Δx = The displacement in the x-direction = 1.0·\hat i

Δy = The displacement in the y-direction = 1.2·\hat j

The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0
3 years ago
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