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trasher [3.6K]
3 years ago
13

You have a great summer job in a research laboratory with a group investigating the possibility of producing power from fusion.

The device being designed confines a hot gas of positively charged ions, called plasma, in a very long cylinder with a radius of 2.0 cm. The charge density of the plasma in the cylinder is 6.0 x 10-5 C/m3. Positively charged Tritium ions are to be injected into the plasma perpendicular to the axis of the cylinder in a direction toward the center of the cylinder. Your job is to determine the speed that a Tritium ion should have when it enters the plasma cylinder so that its velocity is zero when it reaches the axis of the cylinder. Tritium is an isotope of Hydrogen with one proton and two neutrons. You look up the charge of a proton and mass of the tritium in your trusty Physics text to be 1.6 x 10-19 C and 5.0 x 10-27 Kg.
Physics
1 answer:
lozanna [386]3 years ago
3 0

Answer:

The value is v = 2.083 *10^{5} \  m/s

Explanation:

From the question we are told that

The radius of the cylinder is r =  2.0 \ cm  =  0.02 \  m

The charge density of the plasma is J  =  6.0 * 10^{-5} C/m3

Generally the difference in potential energy of the Tritium ion before entering the cylinder and at the axis of the cylinder is mathematically represented as

\Delta U= U_f - U_i  =  \frac{Q *  J  *  r^2}{ 4 *  \epsilon_o}

Here

\epsilon_o} is the permitivity of free space with value

\epsilon_o} =8.85*10^{-12} \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

So

\Delta U  =  \frac{1.60 *10^{-19} *  6.0 * 10^{-5} *  0.02^2}{ 4 *  8.85*10^{-12}}

\Delta U  =  1.085 *10^{-16} \  J

Generally from energy conservation we have that

KE_i + U_i = KE_f + U_f

Given that the velocity at the axis of the cylinder should be zero then

it means that the kinetic energy at this axis will also be zero

So

KE_i + U_i = 0 + U_f

Here KE_i is the initial kinetic energy of the Tritium ion which is mathematically represented as

KE_i  =  \frac{1}{2} *  m * v^2

So

=> \frac{1}{2} *  m * v^2 =  U_f  - U_i

=> \frac{1}{2} *  m * v^2 = \Delta U

=> \frac{1}{2} *  m * v^2 = 1.085 *10^{-16}

=> v =\sqrt{\frac{ 1.085 *10^{-16}}{0.5 * m} }

Here m is the mass of Tritium ion which is given in the question

=> v =\sqrt{\frac{ 1.085 *10^{-16}}{0.5 * 5.0 x 10^{-27} } }

=> v = 2.083 *10^{5} \  m/s

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A particle moves at a constant speed of 34 m/s in a circular path of radius 6.3 m. From this information, what can be calculated
Yakvenalex [24]

Answer:

Centripetal acceleration

Explanation:

  • The centripetal acceleration is the motion inwards towards the center of a circular path.
  • <em><u>Centripetal acceleration is given by; the square of the velocity, divided by the radius of the circular path. </u></em>
  • That is;

         ac = v²/r

         Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the  radius, m

3 0
3 years ago
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
alekssr [168]

Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

Where R_o is the initial resistance

Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

           I = \frac{V}{R}

This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

        \Delta T  = 1795 K

6 0
3 years ago
Most atoms in the Sun exist as:<br><br>A)solids<br>B)liquids<br>C)plasma
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U did it the way i asked. nice lol. its plasma btw
5 0
3 years ago
Read 2 more answers
Question 17 A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off
ivanzaharov [21]

Complete Question:

A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.

Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits

(Question attached)

Answer:

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C (rounded to 1 decimal place)

Explanation:

A calorimeter is used to measure the heat of chemical or physical reactions. The example given in the question is using the calorimeter to determine the specific heat capacity of iron.

When the system reaches equilibrium the iron and water will be the same temperature, T_{e}. The energy lost from the iron will be equal to the energy gained by the water. It is assumed that the only heat exchange is between the iron and water and no exchange with the surroundings.

Q=mc(T_{e}-T_{initial}) (Eq 1)

-Q_{iron}=Q_{water} (Eq 2)

Water:

m_{water}=100.0 g, c_{water}=4.186 J/kg.\°C, T_{initial,water}=23 \°C, T_{e}=27.6 \°C

Iron:

m_{iron}=59.1 g, c_{iron} = ? J/kg.\°C, T_{initial,iron}=85 \°C, T_{e}=27.6 \°C

Substituting Eq 1 into Eq 2 and details extracted from the question:

-m_{iron}c_{iron}(T_{iron,e}-T_{initial})=m_{water}c_{water}(T_{water,e}-T_{initial})

-59.1*c_{iron}(27.6-85)=100.0*4.186(27.6-23)

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C

4 0
3 years ago
A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is comp
Tanzania [10]

Answer:

241 kPa

Explanation:

The ideal gas law states that:

pV=nRT

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

\frac{pV}{T}=nR

For a fixed amount of gas, n is constant, so we can write

\frac{pV}{T}=const.

Therefore, for a gas which undergoes a transformation we have

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K

So we can solve the formula for p_2, the final pressure:

p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa

4 0
3 years ago
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