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1 year ago

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an electron accelerated from rest through a voltage of 770 v enters a region of constant magnetic field. part a if the electron

follows a circular path with a radius of 18 cm , what is the magnitude of the magnetic field?

1 answer:

finlep [7]1 year ago

3 0

The magnitude of the magnetic field on which an electron accelerated from rest through a voltage of 770v enters a region of constant magnetic field is 3.744 * $10^{-4}$ T

From the conservation of energy, we have

$\frac{1}{2}mv^{2} =qV$

$v=\sqrt{\frac{2qV}{m} }$

$=\sqrt{\frac{2*(1,6 * 10^{-19})(770) }{9.11 * 10^{-31} } }$

= $1.644*10^{7$ m/s

At Equilibrium, Centripetal force = magnetic force

$\frac{mv^{2} }{r}=Bqv$

B=mv/rq

by putting, m=9.11* 10^-31 kg

v=1.664*10^7m/s

r=25*10^-2m

q=1.6*10^-19 C

We get, B=3.774* 10^-4 T

Hence the magnitude of the magnetic field, B=3.774* 10^-4 T

Learn more about Magnetic Field here:

brainly.com/question/23096032

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Answer:

$v^2=v_o^2-2\times a\times (v_o.t)$

Explanation:

Given:

Initial velocity of the vehicle, $v_o$

distance between the car and the tree, $x_f$

time taken to respond to the situation, $t$

acceleration of the car after braking, $a$

Using equation of motion:

$v^2=u^2+2a.s$ ..............(1)

where:

$v=$ final velocity of the car when it hits the tree

$u=$ initial velocity of the car when the tree falls

$a=$ acceleration after the brakes are applied

$s=$ distance between the tree and the car after the brakes are applied.

$s=v_o\times t$

Now for this situation the eq. (1) becomes:

$v^2=v_o^2-2\times a\times (v_o.t)$ (negative sign is for the deceleration after the brake is applied to the car.)

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3 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

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Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

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4 years ago

A 20-volt relay has a coil resistance of 200 ohms. How much current does it draw? (I=0.1A)

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Answer:

Explanation:

a) According to ohm's law

V = IR

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I = 20/200

I = 0.1Amperes

b) Using the ohm's law formula

V= IR

Where voltage = 12volts

Current I = 3A

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I = P/V

Given Power = 2Watts

V = 1.5volts

I = 2/1.5

I = 1.33A

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V = P/I

Given P = 90watts

I = 4.5A

V = 90/4.5

V = 20volts

e) Given power = 1.5kW = 1500watts

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3 years ago

An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t

scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

$C = \frac{Q}{V} = e_0\frac{A}{d}$

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

$V = \frac{Q}{e_0} *\frac{d}{A}$

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

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So, if we double the distance between the plates, the potential difference will also be doubled.

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3 years ago

A car bounces up and down on its springs at 1.0 Hz with only the driver in the car. Now the driver is joined by four friends. Th

romanna [79]

The new frequency of oscillation when the car bounces on its springs is 0.447 Hz

<h3>Frequency of oscillation of spring</h3>

The frequency of oscillation of the spring is given by f = (1/2π)√(k/m) where

k = spring constant and
m = mass on spring

Now since k is constant, and f ∝ 1/√m.

So, we have f₂/f₁ = √(m₁/m₂) where

f₁ = initial frequency of spring = 1.0 Hz,
m₁ = mass of driver,
f₂ = final frequency of spring and
m₂ = mass on spring when driver is joined by 4 friends = 5m₁

So, making f₂ subject of the formula, we have

f₂ = [√(m₁/m₂)]f₁

Substituting the values of the variables into the equation, we have

f₂ = [√(m₁/m₂)]f₁

f₂ = [√(m₁/5m₁)]1.0 Hz

f₂ = [√(1/5)]1.0 Hz

f₂ = 1.0 Hz/√5

f₂ = 1.0 Hz/2.236

f₂ = 0.447 Hz

So, the new frequency of oscillation when the car bounces on its springs is 0.447 Hz

Learn more about frequency of oscillation of spring here:

brainly.com/question/15318845

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2 years ago