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3 years ago

Mark all the elementary particlesa) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino

Physics

1 answer:

Natali [406]3 years ago

5 0

Answer:

(a) Proton, (b) electron, (d) Guon, (e) Tau neutrino.

Explanation:

Elementary particles are the particles without any sub strucrure which means they are not composed of other particles.

The elementary particles are classified in three categories which are discussed below:

(1) Quarks: up, down, top, bottom, strange, and charm.

(2) Leptons: electrons, electron neutrino, muon, muon neutrino, tau, tau neutrino.

(3) Bosons: Guon, photons Z bosons, W bosons, higgs.

Also proton consists of up and down quarks so it also consider elementary particles.

Therefore, electron, Guon, proton, and tau neutrino are classified as elementary particles.

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3 years ago

Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the hear

olchik [2.2K]

Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

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i(t) = C(dv/dt)

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3 years ago

If a soccer ball takes 20 seconds to roll 10 meters, what is the average speed of the soccer ball?

xxTIMURxx [149]

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3 years ago

A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

telo118 [61]

Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

$\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}$ (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

$\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}$ (2)

dB/dt = -0.280T/s (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

$E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}$

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

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4 years ago