All categories

2 years ago

??????????????????????????????

Physics

2 answers:

Svet_ta [14]2 years ago

7 0

Answer:

........

Explanation:

Tanzania [10]2 years ago

4 0

Answer:

A) option charging by induction

You might be interested in

Distance is 80 seconds is 8 what is the average speed?

True [87]

Answer:

speed =distance /time

speed =80/8

speed =10m/s

8 0

2 years ago

How does the matter and (solar) energy move through the sphere?

Tema [17]

When the Sun's energy moves through space, it reaches Earth's atmosphere and finally the surface. This radiant solar energy warms the atmosphere and becomes heat energy. This heat energy is transferred throughout the planet's systems in three ways: by radiation, conduction, and convection.

5 0

2 years ago

Which item(s) would be sufficient to make a circuit?

masya89 [10]

Explanation :

A circuit is the representation of the path of the flow of current. The circuit can be either closed or open.

When the switch is off the circuit is closed circuit and when the switch is not connected the circuit is open.

The items that are sufficient to make a circuit are as follows :

Voltage source like a battery.
Resistors or electrical equipment like heater, motor etc.

Other components can be ammeter, voltmeter, ac source, variable resistors etc.

7 0

2 years ago

Read 2 more answers

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

2 years ago

At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi

Nataliya [291]

Answer:

The distance from the charge is 3.35 m.

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

8 0

2 years ago