15 cm

TheE-field strength is 50,000 N/C inside a parallel plate capacitor with 2.0 mmspacing. A proton is released from rest at the po

Ghella [55]

Answer:

The speed is $138852.4 \frac{m}{s}$

Explanation:

Electric field magnitude (E) and electric force magnitude (F) are related by the equation

$F=Eq$

with q the charge of the proton ( $1.61\times10^{-19} C$ ). Because between parallel plates electric field is almost constant, electric force is constant too:

$F=(50000)(1.61\times10^{-19})=8.05\times10^{-15}N$

because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:

$a=\frac{F}{m}$

with m the mass of the proton ( $1.67\times10^{-27} kg$ ):

$a=\frac{8.05\times10^{-15}}{1.67\times10^{-27}}= 4.82\times10^{12}\frac{m}{s^{2}}$

Now with this constant acceleration we can use the kinematic equation

$v^{2}=v_{0}^{2}+2ad$

with v the final speed, $v_i$ the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:

$v=\sqrt{2ad}=\sqrt{2(4.82\times10^{12}\frac{m}{s^{2}})(2.0\times10^{-3}m)}$

$v=138852.4 \frac{m}{s}$

5 0

3 years ago

An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce

ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

$v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}$

From them we can get $v^2=v_0^2+2ad$ .

We have:

$v-v_0=at\\t=\frac{v-v_0}{a}$

And substitute:

$x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}$

We multiply both sides by 2a, and continue:

$2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2$

Being d the displacement $x-x_0$ , we have $v^2=v_0^2+2ad$

For our exercise, we will write this as:

$d=\frac{v^2-v_0^2}{2a}$

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

$d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m$

For the time we use:

$t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s$

6 0

3 years ago

Waves begin to "feel bottom" when the depth of water is

LekaFEV [45]

I think the correct answer would be one half the wavelength. Waves would "feel bottom" when the water is at the depth of 0.5 of the wavelength. "Feel bottom" is a term used to describe that the depth of water affects the wave properties. Hope this answers the question.

8 0

3 years ago

Read 2 more answers

How much power does a hair dryer require to transform 198 000 J of energy in 15 mins?

Marina86 [1]

(198,000 joule / 15 minute) x (minute / 60 sec) = 220 joule/sec = <em>220 watts

</em>

7 0

3 years ago

In a first order decomposition in which the rate constant is 0.204 sec-1, how long will it take (in seconds) until 0.399 mol/L o

DENIUS [597]

Answer:

2.902 to 3 decimal places

5 0

3 years ago