Physics College

An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
sec. What is the average speed of the
electron during this time?

Answers

Answer 1

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

Average speed = 10,000 m/s

Therefore, the average speed of the

electron is 10,000 meters per seconds.

Related Questions

Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone of frequency f, at a speed of 10 m/s directly towards Jason, but his aim is a bit off. As a result, Jason runs forward towards the speaker at a speed of 6 m/s before catching it. Then, the frequency that Jason hears while running can be written as (m/n)f Hz, where m and n are relatively prime positive integers. Compute m n.

Answers

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at  6 m/s .

apparent frequency = [tex]f_o\times\frac{V+v_o}{ V-v_s}[/tex]

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency =  [tex]f\times \frac{340+6}{340-10}[/tex]

= [tex]f\times \frac{346}{330}[/tex]

So m = 346 , n = 330 .

A hammer strikes a nail with a 10 N force for 0.01 seconds. Calculate the impulse of the hammer.

Answers

Answer:

0.1Ns

Explanation:

Impulse is the product of Force and time

Impulse = Force * Time

Given

Force = 10N

Time = 0.01s

Substitute into the formula

Impulse = 10 * 0.01

Impulse = 10 * 1/100

Impulse = 10/100

Impulse = 0.1Ns

hence the impulse of the hammer is 0.1Ns

A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 1.3 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise

Answers

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁)  where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ =  (v² - u²)/2a

h₂ =  h₁ + (v² - u²)/2a

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

h₂ =  1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ =  1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ =  1.3 m + 0.54 m

h₂ =  1.84 m

A 22.0 kg child is riding a playground merry-go- round that is rotating at 40.0 rev/min. What centripetal force must

Answers

Answer:

F = 482.51 N

Explanation:

Given that,

Mass of a child, m = 22 kg

Angular velocity of the merry-go-round, [tex]\omega=40\ rev/min[/tex]

Let the radius of the path, r = 1.25 m

We need to find the centripetal force acting on the child. The formula for the centripetal force is given by :

[tex]F=m\omega^2r\\\\=22\times (4.18879)^2\times 1.25\\\\=482.51\ N[/tex]

So, the required centripetal force is 482.51 N.

Scientists create models to better understand Earth. Which evidence has led scientists to conclude that there are different layers within Earth's interior?

A.analysis of seismic wave data


B.measurement of Earth's diameter


C.temperatures taken within each layer


D.rock samples taken from Earth's core

Answers

Answer:

it is A or D

Explanation:

     

Answer:

ANswer:A

Explanation:

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