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3 years ago

15

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see

the drawing). The plates are separated by a distance of 1.1 cm, and the electric field within the capacitor has a magnitude of 2.2 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

1 answer:

siniylev [52]3 years ago

6 0

Answer:

3.88 * 10^(-15) J

Explanation:

We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.

First, we get the potential and potential energy:

Electric potential = E * r

E = electric field

r = distance between plates

Potential = 2.2 * 10^6 * 0.011

= 2.42 * 10^4 V

The relationship between electric potential and potential energy is:

P. E. = q*V

q = charge of electron = 1.602 * 10^(-19) C

P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)

P. E. = 3.88 * 10^(-15) J

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At what location in a circuit is the electrical potential energy the greatest

Eduardwww [97]

It's not the potential energy. It's just the potential.

It's greatest at the positive terminal of the battery or power supply.

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3 years ago

Suppose a 20-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 15 feet above the ground

Papessa [141]

Answer:

The answer is β=0,85 rads

Explanation:

As the ladder is leaning against the building, we can imagine there´s a triangle where 20ft is the hypotenuse and 15ft is the maximum vertical distance between the ladder and the ground, it means, the leg opposite to β which is the angle we need

Let β(betha) be the angle between the ladder and the ground

We also know that $sin(betha)=(leg opposite)/(hypotenuse)$

In this case we will need to find β, this way:

$betha=sin^-1((15ft/20ft))$

Then β=48,6°

We also have that 2πrads is equal to 360°, in this way we find how much β is in radians:

$betha=(48,6°)*(2pirads/360°)$

then we find β=0,85rads

7 0

3 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0

3 years ago

A person standing a certain distance from four identical loudspeakers is hearing a sound level intensity of 125 dB. What sound l

DENIUS [597]

Answer:

$\mathbf{\beta = 123.75 \ dB}$

Explanation:

From the question, using the expression:

$125 \ dB = 10 \ log (\dfrac{I}{I_o})$

where;

$I_o = 10^{-12} \ W/m^2$

$I = 10^{12.5} \times 10^{-12} \ W/m^2$

$I = 3.162 \ W/m^2$

This is a combined intensity of 4 speakers.

Thus, the intensity of 3 speakers = $\dfrac{3.162\times 3}{4}$

= 2.372 W/m²

Thus;

$\beta = 10 \ log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2$

$\mathbf{\beta = 123.75 \ dB}$

7 0

3 years ago

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

8 0

4 years ago