Ask question

Ask question

All categories

2 years ago

9

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10−2 cm , mounted coaxially within a cylindrical anode

with a radius of 0.5580 cm . The potential difference between the anode and cathode is 400 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.

1 answer:

ch4aika [34]2 years ago

8 0

Answer:

The final speed will be " $1.185\times 10^7 \ m/sec$ ".

Explanation:

The given values are:

Potential difference,

Δv = 400 v

Radius,

r = 0.5580 cm

As we know,

⇒ $W=e \Delta v$

and,

⇒ $\frac{1}{2}mv^2=e \Delta v$

then,

⇒ $v=\sqrt{\frac{2e \Delta v}{m} }$

On substituting the values, we get

⇒ $=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }$

⇒ $=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}$

⇒ $=1.185\times 10^7 \ m/sec$

You might be interested in

How long does it take a wave to travel 1200 meters with the speed of 3 X 108m/sec?

Vladimir79 [104]

Around 3.70 seconds unless traveling through media

5 0

2 years ago

An engine is used to lift a beam weighing 9,800 N up to 145 m. How much work must the engine do to lift this beam? How much work

Arturiano [62]

Explanation:

Given that,

Weight of the engine used to lift a beam, W = 9800 N

Distance, d = 145 m

Work done by the engine to lift the beam is given by :

W = F d

$W=9800\ N\times 145\ m\\\\W=1421000\ J\\\\W=1421\ kJ$

Let W' is the work must be done to lift it 290 m. It is given by :

$W'=9800\ N\times 290\ m\\\\W'=2842000\ J\\\\W'=2842\ kJ$

Hence, this is the required solution.

5 0

3 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

the carnot cycle attempts to model the most efficient possible process by avoiding what? adiabatic processes isothermal processe

kobusy [5.1K]

The carnot cycle attempts to model the most efficient possible process by avoiding irreversible processes.

In essence, the Carnot cycle is a reversible cycle made up of four other reversible processes. A reversible process is one that can be thought of as consisting of a sequence of equilibrium stages because it is carried out endlessly slowly.

Essentially, this means that any reversible cycle can be performed in reverse and that the amount of work or heat exchanged along the forward and backward pathways is the same.

It goes without saying that such reversible processes are not possible because they would take an unlimited amount of time. Therefore, the Carnot Engine is described as an idealized heat engine that uses the Carnot Cycle, a reversible cycle.

Learn more about carnot cycle here;

brainly.com/question/13040188

#SPJ4

5 0

4 months ago

A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold

lubasha [3.4K]

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

fr - Wₓ = m a (1)

Y axis

N- $W_{y}$ = 0

N = W_{y}

let's use trigonometry to find the components of the weight

sin θ = Wₓ / W

cos θ = W_{y} / W

Wₓ = W sin θ

W_{y} = W cos θ

the friction force has the formula

fr = μ N

fr = μ Wy

fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

F = fr

F = (μ mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

4 0

3 years ago