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jek_recluse [69]
3 years ago
6

(1) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy so

urce is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.0 x 10^7 m/s in a circular path 2.00 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.
(2) Is this field strength obtainable with today’s technology or is it a futuristic possibility?
Physics
1 answer:
malfutka [58]3 years ago
7 0

Solution :

The relationship between the strength of magnetic field and the radiusof a charged particle's path is obtained through Newton's second law, which is given by :

F = ma

F = qvB and $a=\frac{v^2}{r}$

Substituting these values in the second law of Newton,

$qvB=\frac{mv^2}{r}$

Now solving for B, we get:

$B = \frac{mv}{rq}$

  $=\frac{(1.67 \times 10^{-27})(5 \times 10^{7})}{2\times 1.6 \times 10^{-19}}$

 = 0.261 T

The field strength can be obtained by using the technology of today.

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Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
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  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

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6 0
3 years ago
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Answer:

1.been both -ve charged or both +be charged particles

2. 3.52mC

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For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.

Now the Force sustain by the extended string is

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Where K is the force constant of the string, 320 N/m

e is the extension,0.033 m

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F = kQ1 Q2

But Q1=Q2{ since the charge are of the same magnitude}.

Hence F = KQ^2

Where K is columns constant =9×10^9F/m

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Q= √10.56/9×10^9

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