4. What is the period of the microwaves in the above question?

Physics

1 answer:

Sidana [21]2 years ago

6 0

Answer:

The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle.

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A proton starts from rest near the surface of a sheet of charge. It experiences a force of 14.0 microNewtons towards the sheet.

Naddika [18.5K]

Answer:

The surface charge density of the sheet is 1547.516 C/m²

Explanation:

Given;

force experienced by the proton, F = 14.0 μN

Electric field near an infinite sheet with surface charge density (σ) is given as;

E = σ/2ε₀

The force of attraction the proton experienced towards the sheet;

F = Eq

F = q(σ/2ε₀) $= \frac{q \sigma}{2 \epsilon_o}$

where;

q is the charge of the proton

σ is the surface charge density of the sheet

ε₀ is the permittivity of free space

$F = \frac{q \sigma}{2 \epsilon_o} \\\\\sigma = \frac{2F \epsilon_o}{q} \\\\\sigma = \frac{2*14*10^{-6}*8.854*10^{-12}}{1.602*10^{-19}} = 1547.516 \ C/m^2$

Therefore, the surface charge density of the sheet is 1547.516 C/m²

6 0

2 years ago

The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current

Komok [63]

Answer:

$I=0.047A$

Explanation:

Let's use Ohm's law:

$V=IR$

$I=\frac{V}{R}$ (1)

Where:

$V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance$

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

$R=\rho*\frac{l}{A}$ (2)

Where:

$R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}$

Keep in mind that the electrical resistivity of the gold is a known constant which is $\rho_g_o_l_d=2.35*10^{-8}$ and the cross sectional area of the conductor is calculated as: