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snow_tiger [21]
3 years ago
8

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

Physics
1 answer:
Dominik [7]3 years ago
5 0

a) 0.261 T

b) this field strength is obtainable with today's technology

Explanation:

a)

The force experienced by a charged particle moving perpendicular to a magnetic field is given by

F=qvB

where

q is the charge

v is the velocity

B is the strength of the field

This force is perpendicular to the motion of the particle, which therefore moves in a circular path; and so, this force acts as centripetal force, so we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the circle

In this problem, we have:

q=1.6\cdot 10^{-19}C (magnitude of the charge of antiprotons)

v=5.00 \cdot 10^7 m/s (velocity)

m=1.67\cdot 10^{-27}kg (mass of antiprotons)

r = 2.00 m (radius)

Therefore, we can re-arrange the equation and solve to find B, the magnetic field strength:

B=\frac{mv}{qr}=\frac{(1.67\cdot 10^{-27})(5.00\cdot 10^7)}{(1.6\cdot 10^{-19})(2.00)}=0.261 T

B)

The strength of the magnetic field calculated in part A) is

B=0.261 T

This is indeed a very strong magnetic field. In fact, by comparison, the Earth's magnetic field has a strength of about

B_{earth}=5\cdot 10^{-5} T

However, there are current technologies available that are able to produce such strong fields. For instance, the superconducting magnets in the LHC (Large Hadron Collider) are able to produce magnetic fields of strength up to 8 Tesla (8 T).

Therefore, we can say that this field strength is obtainable with today's technology.

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Calculate the momentum of a 1,500 kg car traveling at 6 m/s.
mamaluj [8]

Answer: 9000 kgm/s

Explanation:

Mass of car = 1500 kg

Speed by which car moves = 6 m/s. Momentum of the car = ?

Recall that:

Linear momentum = Mass x Speed

= 1500kg x 6m/s

= 9000 kgm/s

Thus, the linear momentum of the car is 9000 kgm/s

3 0
3 years ago
In order to transmit information via radio waves, the waves need to be changed somehow. For car radios this can happen in two wa
laila [671]

Answer: amplitude

Explanation: This describes the maximum amount of the displacement of a particle from it rest position. Usually, it is measured in metres

Since we are considering AM which is amplitude modulation, a technique used in electronic communication, most commonly for broadcasting information through a radio carrier wave. In amplitude modulation, the amplitude (signal strength) of the carrier wave is diversified in proportion to that of the message signal being broadcasted.

7 0
3 years ago
Russell bradley carried 207 kg of bricks 3.65 m up a ladder. If the amount of work required to perform that task is used to comp
MrRa [10]

Answer:

Explanation:

Work done in carrying bricks

mgh

= 207 x 9.8 x 3.65

-= 7404.4 J

Work done in compressing gas

PΔV

Pressure x change in volume

1.8 x 10⁶ ΔV = 7404.4

ΔV  = 7404.4  / 1.8 x 10⁶m³

= 4113.33 x 10⁻⁶ m³

= 4113.33 cc

8 0
3 years ago
Read 2 more answers
Please explain how to solve 17 and 18
andriy [413]

Answer:

17. h = l − l cos θ

18. 1.40 m

Explanation:

Let's call d the height of the triangle.  We can then say:

h = l − d

Using trig, we can write d in terms of l and θ:

d = l cos θ

h = l − l cos θ

If l = 6 m and l cos θ = 40°:

h = 6 − 6 cos 40

h ≈ 1.40

3 0
3 years ago
Two charges are located in the x-y plane. If q1=-4.55 nC and is located at x=0.00 m, y=0.680 m and the second charge has magnitu
Elden [556K]

Answer:

Ex= -23.8 N/C  Ey = 74.3 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1, as follows:

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:

E₁ = k*(4.55 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:

r₂² = 1²m² + (0.6)²m² = 1.36 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855

By the same token, sin θ can be obtained as follows:

sin θ = - (0.6 m / 1.17 m) = -0.513

⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)

⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  88.6 N/C + (-14.3 N/C) =74.3 N/C

7 0
3 years ago
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