Ask question

Ask question

All categories

3 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]3 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

You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The

Vinil7 [7]

The work done is 8427 J while the energy expended is 80.25 W.

<h3>What is work done?</h3>

Work done is defined as the product of the force and the distance. We know that the work done in a gravitational field is given as;

W = mgh

Total mass of the water bucket and chain = 13.9 kg + 19.3 kg = 33.2Kg

Distance covered = 25.9 m

W = 33.2Kg * 9.8 m/s^2 * 25.9 m

W = 8427 J

Recall that the work done = Energy expended

Power = Energy expended/ Time

Power = 8427 J/ 1.75 * 60 seconds

Power = 80.25 W

Learn more about work and energy:brainly.com/question/17290830

#SPJ1

5 0

2 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

frez [133]

Answer:

<em>1.49 x </em> $10^{11}$ <em></em>

<em></em>

Explanation:

Kepler's third law states that <em>The square of the orbital period of a planet is directly proportional to the cube of its orbit.</em>

Mathematically, this can be stated as

$T^{2}$ ∝ $R^{3}$

<em>to remove the proportionality sign we introduce a constant</em>

$T^{2}$ = k $R^{3}$

k = $\frac{T^{2} }{R^{3} }$

Where T is the orbital period,

and R is the orbit around the sun.

For mars,

T = 687 days

R = 2.279 x $10^{11}$

for mars, constant k will be

k = $\frac{687^{2} }{(2.279*10^{11}) ^{3} }$ = 3.987 x $10^{-29}$

For Earth, orbital period T is 365 days, therefore

$365^{2}$ = 3.987 x $10^{-29}$ x $R^{3}$

$R^{3}$ = 3.34 x $10^{33}$

R =<em> 1.49 x </em> $10^{11}$ <em></em>

8 0

3 years ago

Read 2 more answers

The figure in Figure 1 shows two single-slit diffraction patterns. The distance between the slit and the viewing screen is the s

V125BC [204]

Answer:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."

Explanation:

The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.

D sin θ = m λ

if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."

This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.

Hope this helps!

8 0

2 years ago

Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v

wlad13 [49]

1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call $v_f$ the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
$m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f$ (1)
where
$m_A=7 kg$ is the mass of ball A
$m_B=2 kg$ is the mass of ball B
$v_{Ai}=6 m/s$ is the initial velocity of ball A
$v_{Bi}=-12 m/s$ is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find $v_f$ , we find that the final velocity of the balls is
$v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s$
and the positive sign means the two balls are going to the right.

2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
$m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}$
$\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2$
where
$v_{fA}$ is the final velocity of ball A
$v_{fB}$ is the final velocity of ball B

If we solve simultaneously the two equations, we find:
$v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s$
$v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s$
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
$p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s$
and the total momentum after the collision is:
$p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s$

3 0

3 years ago

Is abolute zero equal to 0c

Tamiku [17]

No silly. Absolute zero is -273.1 C.

6 0

3 years ago

Read 2 more answers