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2 years ago

12

What is the formula for time? (something like speed *distance=time??)?

1 answer:

butalik [34]2 years ago

4 0

It would be: Time = Distance / Speed

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A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th

GREYUIT [131]

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

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

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

$v=\frac{qBr}{m}$

where in this problem we have:

$q=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$B=208 G=208\cdot 10^{-4}T$ is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$

6 0

3 years ago

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago

Hii please help i’ll give brainliest if you give a correct answer please please hurry

fiasKO [112]

Answer:

B most likely

Explanation:

tell me if this is right or wrong

8 0

2 years ago

A one bucket is fill up with sand and another is empty .Which bucket can oscilliate more when release from the distance

Sav [38]

Answer:

a one bucket with full of sand because of it's mass which plays role in momentum

6 0

2 years ago

A jet airliner moving initially at 548 mph

sasho [114]

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

$v_1 =548 mph$ along the x-direction

$v_2 = 343 mph$ in a direction $67^{\circ}$ north of east.

To find the resultant, we must resolve both vectors on the x- and y- axis:

$v_{1x}= 548 mph$

$v_{1y}=0$

$v_{2x} = (343 mph)( cos 67^{\circ})=134.0 mph$

$v_{2y} = (343 mph)( sin 67^{\circ})=315.7 mph$

So, the components of the resultant velocity in the two directions are

$v_{x}=548 mph+134 mph=682 mph$

$v_{y}=0 mph+315.7 mph=315.7 mph$

So the new speed of the aircraft is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(682 mph)^2+(315.7 mph)^2}=751.5 mph$

3 0

3 years ago