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Lena [83]
4 years ago
6

Two light bulbs are being used in a lab experiment with a single battery. Bulb A has half the resistance as Bulb B. Which bulb i

s the dimmest in series? Which bulb is the brightest in parallel? Explain.
Physics
1 answer:
Naya [18.7K]4 years ago
3 0

Answer and Explanation:

It is given that bulb A has the half the resistance as B

WHEN THEY ARE CONNECTED IN SERIES : In series circuit we know that current is same in the whole circuit

The power is given by P=i^2R the expression it is clear that the bulb which has more resistance consume more power as current is same, and the bulb which consume more power will be brightest

It is given that resistance of bulb A is half the resistance of B So bulb A will be dimmest in this case

WHEN THEY ARE CONNECTED IN PARALLEL : In parallel circuit voltage is same across the circuit

The power is given by P=\frac{V^2}{R} from the expression it is clear that the bulb which has least resistance will consume more power and so will be brightest

As the resistance of bulb A is less so it will be brightest in this case

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In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
MA_775_DIABLO [31]

A) Francium-223

In an alpha decay, a nucleus decay emitting an alpha particle, which corresponds to a nucleus of helium: so, it consists of 2 protons and 2 neutrons.

X \rightarrow X' + \alpha

This means that in the decay:

- The original nucleus loses 2 protons --> so its atomic number Z decreases by 2 units

- The original nucleus loses 2 nucleons (2 protons and 2 neutrons) --> so its mass number A decreases by 4 units

In this example, the original nucleus is Ac (Actinium), with

Z = 89

A = 227

After the decay, it must be

Z - 2 = 89 - 2 = 87

A - 4 = 227 - 4 = 223

We see from the periodict table, Z=87 corresponds to Francium (Fr), so the final nucleus will be francium-223 (the isotope of francium with 223 nucleons).

B) Polonium-211

In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.

n \rightarrow p + e^- + \bar{\nu}

Therefore, in this process:

- The original nucleus gains 1 protons, so its atomic number Z increases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Bi (bismuth)-211, with

Z = 83

A = 211

So After the decay, it will be

Z + 1 = 83 + 1 = 84

A = 211

So, the nucleus will be Polonium (Z=84), isotope with 211 nucleons.

C) Neon-22

In a beta-plus decay, a proton in the nucleus turns into a neutron, emitting a fast-moving positron (the beta particle) and a neutrino.

p \rightarrow n + e^+ +\nu

Therefore, in this process:

- The original nucleus loses 1 protons, so its atomic number Z decreases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Na (sodium)-22, with

Z = 11

A = 22

So After the decay, it will be

Z - 1 = 11 - 1 = 10

A = 22

So, the nucleus will be Neon (Z=10), isotope with 22 nucleons.

D) Technetium-98

In a gamma decay, an unstable nucleus emits a gamma ray:

X' \rightarrow X + \gamma

In this process, only energy is released (in the form of gamma ray), so there is no gain/loss of protons/neutrons in the process. This means that:

- The atomic number Z remains constant

- The mass number A remains constant

In this example, we have a nucleus of Tc (Technetium)-98, with

Z = 43

A = 98

These numbers will not change during the decay: this means that after the decay, we will still have a nucleus of Technetium-98.

8 0
3 years ago
A high-temperature, gas-cooled nuclear reactor consists of a composite, cylindrical wall for which a thorium fuel element (kth =
WARRIOR [948]

Answer:

a) T_1 = 938 K , T_2 = 931 K

b) To prevent softening of the materials, which would occur below their  melting points, the reactor should not be operated much above:

                                      q = 3*10^8 W/m^3

Explanation:

Given:

- See the attachment for the figure for this question.

- Melting point of Thorium T_th = 2000 K

- Melting point of Thorium T_g = 2300 K

Find:

a) If the thermal energy is uniformly generated in the fuel element at a rate q = 10^8 W/m^3 then what are the temperatures T_1 and T_2 at the inner and outer surfaces, respectively, of the fuel element?

b) Compute and plot the temperature distribution in the composite wall for selected values of q.  What is the maximum allowable value of q.

Solution:

part a)

- The outer surface temperature of the fuel, T_2, may be determined from the rate equation:

                                 q*A_th = T_2 - T_inf / R'_total

Where,

           A_th: Area of the thorium section

           T_inf: The temperature of coolant = 600 K

           R'_total: The resistance per unit length.

- Calculate the resistance per unit length R' from thorium surface to coolant:

           R'_total = Ln(r_3/r_2) / 2*pi*k_g + 1 / 2*pi*r_3*h

Plug in values:

           R'_total = Ln(14/11) / 2*pi*3 + 1 / 2*pi*0.014*2000

           R'_total = 0.0185 mK / W

- And the heat rate per unit length may be determined by applying an energy balance to a control surface  about the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that:

           q' = q*A_th = q*pi*(r_2^2 - r_1^2)

           q' = 10^8*pi*(0.011^2 - 0.008^2) = 17,907 W / m

Hence,

           T_2 = q' * R'_total + T_inf

           T_2 = 17,907*0.0185 + 600

          T_2 = 931 K

- With zero heat flux at the inner surface of the fuel element, We will apply the derived results for boundary conditions as follows:

 T_1 = T_2 + (q*r_2^2/4*k_th)*( 1 - (r_1/r_2)^2) - (q*r_1^2/2*k_th)*Ln(r_2/r_1)

Plug values in:

 T_1 = 931+(10^8*0.011^2/4*57)*( 1 - (.8/1.1)^2) - (10^8*0.008^2/2*57)*Ln(1.1/.8)

 T_1 = 931 + 25 - 18 = 938 K

part b)

The temperature distributions may be obtained by using the IHT model for one-dimensional, steady state conduction in a hollow tube. For the fuel element (q > 0),  an adiabatic surface condition is  prescribed at r_1 while heat transfer from the outer surface at r_2 to the coolant is governed by the thermal  resistance:

                              R"_total = 2*pi*r_2*R'_total

                              R"_total = 2*pi*0.011*0.0185 = 0.00128 m^2K/W

- For the graphite ( q = 0 ), the value of T_2 obtained from the foregoing solution is prescribed as an inner boundary condition at r_2, while a convection condition is prescribed at the outer surface (r_3).

- For 5*10^8 < q and q > 5*10^8, the distributions are given in attachment.

The graphs obtained:

- The comparatively large value of k_t yields small temperature variations across the fuel element,  while the small value of k_g results in large temperature variations across the graphite.

Operation  at q = 5*10^8 W/^3  is clearly unacceptable, since the melting points of thorium and graphite are exceeded  and approached, respectively. To prevent softening of the materials, which would occur below their  melting points, the reactor should not be operated much above:

                                      q = 3*10^8 W/m^3

6 0
3 years ago
A skater increases her velocity from 2.0 m/sec to 10.0 m/sec in 3.0 sec. what is the acceleration of the skater
Kaylis [27]

Final speed 10.0 rn/s Change in time = 3 seconds The acceleration of the skater is 2.7 meters per second Relationship per second. 3.

3 0
3 years ago
What is the lowest level of sound that can be detected by human hearing? 0 Hz 10 Hz 20 Hz 100 Hz
maxonik [38]
These choices are not "levels" of sound.  They're frequencies of sound.

The lower limit of audible frequency is usually considered to be 20 Hz.
But there are some people with ears that can hear 10 Hz, and others ...
particularly the elderly ... who can't hear anything below 100 Hz.

The question is a rather poor one.
8 0
4 years ago
Read 2 more answers
Help
malfutka [58]
The answer for this is 1200N
6 0
3 years ago
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