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morpeh [17]
3 years ago
11

A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p

roton. (A) Find the speed of the proton. (B) Find the magnitude of magnetic force on the proton.
Physics
1 answer:
PIT_PIT [208]3 years ago
6 0

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

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A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication
koban [17]

Answer:

r = 4.24x10⁴ km.  

     

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0
3 years ago
An electric toy with a resistance of 2.50 Ω is operated by a 3.00-V battery. (a) What current does the toy draw? (b) Assuming th
Leno4ka [110]

Answer:

a) The current is i = 1.2 A

b) The charge is Q = 17280 C

c) The energy is E = 43200 J

Explanation:

a) The current is given by the ohm's law wich is:

i = V/R = 3/2.5 = 1.2 A

b) Since the charge is steady we can use the following equation to find the charge amount in that time:

i = Q/t

Q = t*i

Where t is in seconds, so we have 4h * 3600 = 14400 s

Q = 1.2*14400 = 17280 C

c)  The energy is the power delivered to the toy multiplied by the time:

P = 1.2*2.5 = 3 W

E = P*t = 3*14400 = 43200 J

7 0
3 years ago
A horizontal uniform meter stick supported at the 50-cm mark has a mass of 0.50 kg hanging from it at the 20-cm mark and a 0.30
ElenaW [278]

Answer:

70 cm

Explanation:

0.5 kg at 20 cm

0.3 kg at 60 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

0.5(50-20)=0.3(60-50)+0.6x\\\Rightarrow x=\dfrac{0.5(50-20)-0.3(60-50)}{0.6}\\\Rightarrow x=20\ cm

The position of the third mass of 0.6 kg is at 20+50 = 70 cm

7 0
3 years ago
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Assume that the same amount of current passes through the electromagnets shown. Identify which electromagnet is stronger than th
hodyreva [135]

First electromagnet

Explanation:

The first electromagnet is the strongest and it is stronger than the given electromagnet above.

An electromagnet is a temporary magnet made by passing current through a wire wounded round an iron core or metallic core.

  • When current is passed through, the metal becomes magnetic.
  • The strength of the electromagnet depends on the number of coil round the metal core and also the intensity of current passed through it.
  • The higher the number of coils wounded round the metal core, the stronger the electromagnet that will be produced.
  • Also, the higher the intensity of electricity passed through the wire, the stronger it is.

learn more:

Electromagnet brainly.com/question/2191993

#learnwithBrainly

3 0
3 years ago
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Scientists in a test lab are testing the hardness of a surface before constructing a building. Calculations indicate that the en
densk [106]
From the given problem, a limit on the depression of a building is placed at 20 centimeters. To solve how many floors can be safely added, a quantity of how many cm will a building sink for each floor that is added is needed. Unfortunately, it is not found anywhere in the problem. However, we can provide a formula to solve for the depression. This is as follows:

Building depression < 20 cm

Building depression = (cm depression per floor) * (no. of floors)
7 0
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