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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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The safest way to view an image of the sun is to use.
Nostrana [21]

Answer:

Pinhole Projection

Explanation:

You will get a great view if you project a sunlight coming thrugh a small hole onto a viewing card enclosed in a long box. The smaller the pinhole, the sharper the image

7 0
2 years ago
a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100
Anastaziya [24]

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

       

let's reduce the magnitudes to the SI system

         r = 10 cm = 0.10 m

         m = 10 g = 0.010 kg

         A = 100 cm² = 0.01 m²

the definition of density is

          ρ = m / V

the volume of a sphere

         V = \frac{4}{3} \ \pi r^{3}

          V = \frac{4}{3} π 0.1³

          V = 4.189 10⁻³ m³

let's calculate the density of the ball

           ρ = \frac{0.010}{4.189 \ 10^{-3} }

           ρ = 2.387 kg / m³

the tabulated density of water is

         ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

            B - W = 0

            B = W

             

where B is the thrust that is given by Archimedes' principle

           ρ_liquid  g V_submerged = m g

           V_submerged = m / ρ_liquid

we calculate

            V _submerged = 0.10 9.8 / 997

             V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

           V = A h

            h = V / A

let's calculate

            h = 9.83 10⁻⁴ / 0.01

            h = 0.0983  m

this is equal to h = 9.83 cm

8 0
2 years ago
the mass of one water drop is 0.0008kg and the gravitational field strength is 10N/kg what is its weight
djyliett [7]
Weight = (mass) x (gravity)

Weight = (8 x 10⁻⁴ kg) x (10 N/kg) = 0.008 Newton
8 0
3 years ago
The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo
hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

W-N=0 (1)

the weight of the hyppo is

W=mg=(1500 kg)(9.8 m/s^2)=14,700 N

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

N=W=14,700 N

8 0
3 years ago
A spelunker is surveying a cave. She follows a passage 140 m straight west, then 290 m in a direction 45? east of south, and the
zimovet [89]

Answer:

Magnitude of fourth displacement is approximately 95 metres,

Direction of fourth displacement is straight west.

3 0
3 years ago
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