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3 years ago

What happens when you stroke the prongs with the wire?

1 answer:

8 0

When you stroke the prongs with a wire, a static sound is being produced. This kind of minimum-information noise is called white noise, by analogy with white light which is a uniform mixture of all the different possible colors. Hope this helps.

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A ball is held at rest at the top of a hill. The ball is then released and starts rolling down the hill. At the bottom it reache

Alex17521 [72]

Answer:

Gravitational Potential Energy

Explanation:

a ball is held rest at the top of hill

gravitational potential energy will store due to its height

it. and body will start move downward and its potential energy will convert into kinetic energy due to motion of body

at the ground level it will stop and potential energy will became zero and kinetic energy get convert into internal energy due to collisions

3 0

3 years ago

Ethan moves a 100 kg barrel up a 50 meter ramp. The amount of work Ethan has completed is:

Zinaida [17]

Answer:

I think C. 150 Joules

Explanation:

4 0

3 years ago

Read 2 more answers

Dimensional Analysis : 3 days to seconds

LuckyWell [14K]

Hi,

To convert 3 days to seconds write this.

1h = 3600s
24h = 3600 · 24 = 86400
3 days = 3 · 86400 = 259200sec

Hope this helps.
r3t40

8 0

3 years ago

Read 2 more answers

An electron follows a helical path in a uniform magnetic field of magnitude 0.340 T. The pitch of the path is 6.00 µm, and the m

dedylja [7]

Answer:

The electron's speed is 34007.35 m/s

Explanation:

It is given that,

Magnetic field, B = 0.34 T

Magnetic force on the electron, $F=1.85\times 10^{-15}\ N$

The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :

$F=B\times q\times v$

q = charge on an electron, $q=1.6\times 10^{-19}\ C$

v = velocity of an electron

$v=\dfrac{F}{Bq}$

$v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}$

v = 34007.35 m/s

Hence, this is the required solution.

4 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

4 years ago