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2 years ago

Why do you sometimes get shocked by static electricity? Explain how it happens and give an example.

Physics

1 answer:

laiz [17]2 years ago

5 0

Answer:

That tiny shock you feel is a result of the quick movement of these electrons. You can think of a shock as a river of millions of electrons flying through the air. ... Static electricity happens more often during the colder seasons because the air is drier, and it's easier to build up electrons on the skin's surface.

Explanation:

Don't copy excatly but here it is.

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In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

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

So, $v = \sqrt{\frac{2E}{m}}$ .... (1)

Now,

$r = \frac{mv}{Bq}$

Substitute the value of v from equation (1), we get

$r = \frac{\sqrt{2mE}}{Bq}$

Let the radius of the alpha particle is r2.

For proton

So, $r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0

2 years ago

Read 2 more answers

How does an atom develop a positive charge

Dmitry [639]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ A normal atom has the same amount of electrons and protons, making it neutral. An atom develops a positive charge when it loses an electron(s). Once it loses an electron(s), there would now be more protons that electrons.

Short answer: by losing an electron(s)

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

8 0

3 years ago

Jeff’s father is installing a do-it-yourself security system at his house. He needs to get a device from his workshop that conve

elena55 [62]

The answer is a buzzer. hope this helps. this is the answer because it use electricity to make the sound please award brainliest

7 0

2 years ago

Read 2 more answers

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

7 0

2 years ago

Jim is driving a 2268-kg pickup truck at 22 m/s and releases his foot from the accelerator pedal. The car eventually stops due t

shutvik [7]

Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

$E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J$

The work done by the friction force is given by:

$W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m$

6 0

2 years ago