1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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.

You might be interested in
An athlete whirls a 7.00 kg hammer 1.8 m from the axis of rotation in a horizontal circle, as
iogann1982 [59]

Answer:

A-500 N

Explanation:

The computation of the tension in the chain is shown below

As we know that

F = ma

where

F denotes force

m denotes mass = 7

And, a denotes acceleration

Now for the acceleration we have to do the following calculations

The speed (v) of the hammer is

v = Angular speed × radius

where,

Angular seed = 2 × π ÷ Time Period

So, v = 2 × π × r ÷ P

v = 2 × 3.14 × 1.8 ÷ 1

= 11.304 m/s

Now

a = v^2 ÷ r

= 70.98912 m/s^2

Now the tension is  

T = F = m × a

= 7 × 70.98912

= 496.92384 N

= 500 N

5 0
3 years ago
If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 20° with a set of planes in a crystal causing first order constru
Flura [38]

Answer:

43.16°

Explanation:

λ = Wavelength = 1.4×10⁻¹⁰ m

θ₁ = 20°

n can be any integer

d = distance between the two slits

Since for the first bright fringe, n₁ = 1

n₂ = 2 for second order line

The relation between the distance of the slits and the angle through which it is passed is:

dsinθ=nλ

As d and λ are constant

\frac{n_1\lambda}{sin \theta_1}=\frac{n_2\lambda}{sin \theta_2}\\\Rightarrow \frac{1}{sin20}=\frac{2}{sin\theta_2}\\\Rightarrow sin\theta_2=\frac{2}{\frac{1}{sin20}}\\\Rightarrow \theta_2=sin^{-1}{\frac{2}{\frac{1}{sin20}}}\\\Rightarrow \theta_2=43.16^{\circ}

∴ Angle by which the second order line appear is 43.16°

5 0
3 years ago
In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into
NNADVOKAT [17]

Answer:

2) a_y= -g  3) vₓ=constant v_y = v_{oy} - g t, 4)  vₓ = v₀ₓ - ax t

5)  changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

           v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

           vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

           x = v₀ₓ t

           y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

4 0
3 years ago
A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant
Alla [95]

Answer:

a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

Given the data in the question;

mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

specific volume v₁ = 0.5092 m³/kg

so the properties of steam at p₂ = 550 kPa, and dryness fraction

x = 0.4

specific volume v₂ = v_f + xv_{fg

v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )

v₂ = 0.1377 m³/kg

Now, work done during the process;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.1377 - 0.5092 )

W = 5500 × -0.3715

W = -2043.25 kJ

Therefore, the work (W) done during the process is -2043.25 kJ

( The negative, indicates work is done on the system )

b)

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

x₂ = 100% - 80% = 20% = 0.2

specific volume v₂ = v_f + x₂v_{fg

v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )

v₂ = 0.06939 m³/kg

Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.06939 - 0.5092 )

W = 5500 × -0.43981

W = -2418.96 kJ

Therefore, the work (W) done during the process is -2418.96 kJ

3 0
3 years ago
10 solve the following humeric problem - S. a What is the lift height of salu? T. object energy required to 25 kg to mass of a 1
Salsk061 [2.6K]
  • Mass=m=25kg
  • Height=h=10m
  • Acceleration due to gravity=g=10m/s^2

\boxed{\sf E_P=mgh}

\\ \sf\longmapsto E_P=25(10)(10)

\\ \sf\longmapsto E_P=25(100)

\\ \sf\longmapsto E_P=2500J

6 0
3 years ago
Other questions:
  • PLEASE ANSWER QUICK!!! 2. Every magnet has _ unlike poles.
    7·1 answer
  • An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay trav
    7·1 answer
  • In 1-2 sentences describe how your nose cleans the air you inhale.
    15·1 answer
  • A fully loaded, slow-moving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 49 m in
    13·1 answer
  • In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector
    12·1 answer
  • Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.
    12·1 answer
  • Explain why it is not advisable to be in a garage when the car engine is being <br> heated.​
    7·2 answers
  • (b)
    14·1 answer
  • Why are the Soviets ahead in the race to capture the V-2 rocket and Werner von Braun?
    15·1 answer
  • Spiderman is standing on a building and notices some shady characters up to no good on the street below. Rather than leap or
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!