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3 years ago

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The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d

ifference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.
1. What is the electric field strength between the plates?
2. With what speed does an electron exit the electron gun if its entry speed is close to zero?

1 answer:

yulyashka [42]3 years ago

3 0

Answer:

a) F = 3.2 10⁻¹⁰ N , b) v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

F = q E

The electric field is related to the potential reference

V = E d

E = V / d

Let's replace

F = e V / d

Let's calculate

F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

v² = v₀ + 2 a d

v = √ 2 ad

Acceleration can be found with Newton's second law

e V / d = m a

a = e / m V / d

a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

v = √ (98,448 10¹⁴)

v = 9.9 10⁷ m / s

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A man wishes to row the shortest possible distance from north to south across a river which is flowing at 2 km/hr from the east.

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From the diagram we have that

$sin\theta = \frac{2}{4}$

$\theta = sin^{-1} (\frac{1}{2})$

$\theta = 30\°$

Therefore the direction is 30° from east of south

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3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

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

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

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There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. Part A A p

tamaranim1 [39]

Explanation:

A pilot flying a weight; it takes t = 3.4 s to hit the ground, during which it travels a horizontal distance d = 120 m

h = g/2*t^2

h= 4.903*3.4^2

h=56.701 m

V = d/t

V= 120/3.4

V=35.3 m/sec

Now the pilot does a run at the same height but twice the speed.

How much time (t1) does it take the weight to hit the ground?

t1 = t = 3.4 sec (depending just upon height)

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A photographer in a helicopter ascending vertically at a constant rate of accidentally drops a camera out the window when the he

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Answer:

The time it takes for the camera to reach the ground is 5 s.

Explanation:

To solve this problem, we will use the free fall cinematic equation.

Since the helicopter ascends with constant speed, the camera falls to the ground only by the effect of gravity on it.

The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 $\frac{m}{s}$ .

First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.

To calculate the maximum height to which it arrives, we will use the formula of vertical shot (since the camera leaves the helicopter with a speed upwards of 12,5 $\frac{m}{s}$ ).

$V_{f} ^{2}$ = $V_{0} ^{2}$ - 2 * g * h

Where:

$V_{f}$ : final speed at maximum height.

$V_{0}$ : initial speed when it falls from the helicopter.

g: gravity taken at 9.8 $\frac{m}{s^{2} }$

h: height reached from 60 m when leaving the helicopter

as $V_{f}$ =0

0= $(12,5\frac{m}{s}) ^{2}$ - 2 * $9,8\frac{m}{s^{2} }$ * h

clear h:

h= $(12,5 \frac{m}{s^{2} } )^{2}$ / (2 * $9,8 \frac{m}{s^{2} }$ )

h=7,97 m

Then we must calculate the time it takes to reach its maximum height:

$V_{f}$ = $V_{0}$ - g * t

t: time it takes to arrive from the moment it leaves the helicopter at its maximum height.

as $V_{f}$ =0

0=12.5 $\frac{m}{s}$ - 9.8 $\frac{m}{s^{2} }$ * t

clearing t

t=12.5 $\frac{m}{s}$ / 9.8 $\frac{m}{s^{2} }$

t=1.27 s.

Now we can calculate the time it takes to fall from the maximum height of 67.97 m.

The equation we will use is Y= $v_{0}*t+(\frac{g*t^{2} }{2} )$

where:

t: time it takes for the camera to fall.

Y: height from where the camera falls concerning the ground.

$v_{0}$ : initial speed of the camera at the time of starting the fall.

g: acceleration of gravity, estimated at 9.8 $\frac{m}{s^{2} }$

Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.

$v_{0}$ =0

So the first term of our equation is nullified.

Step 2: To calculate the time it takes to fall, we clear "t" of the equation:

Y= $\frac{(g*t^{2})}{2}$

Y*2=(g* $t^{2}$ )

$\frac{Y*2}{g}$ = $t^{2}$

$\sqrt{\frac{Y*2}{g} }$ =t

Step 3: I replace the values with the incognites and get "t".

t= $\sqrt{\frac{67,97m*2}{9,8\frac{m}{s^{2} } } }$

t=3,73 s

The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.

t= 1,27 s + 3,73 s = 5 s

Have a nice day!

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