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2 years ago

What is the wavelength of a wave that is traveling at 343 m/s with a frequency of 1000 Hz

Physics

1 answer:

ZanzabumX [31]2 years ago

3 0

Answer:

$343 \div 1000 = 0.343$

wavelength is equivalent to the velocity divided by the frequency

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Two resistors, of R1 = 3.11 Ω and R2 = 6.15 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna

Firlakuza [10]

Part (a):
1- Since the resistors are in series, therefore, the total resistance is the summation of the two resistors.
Therefore:
Rtotal = R1 + R2 = 3.11 + 6.15 = 9.26 ohm

2- Since the two resistors are in series, therefore, the current flowing in both is the same. We will use ohm's law to get the current as follows:
V = I*R
V is the voltage of the battery = 24 v
I is the current we want to get
R is the total resistance = 9.26 ohm
Therefore:
24 = 9.26*I
I = 24 / 9.26
I = 2.59 ampere

Part (b):
To get the voltage across the second resistor, we will again use Ohm's law as follows:
V = I*R
V is the voltage we want to get
I is the current in the second resistor = 2.59 ampere
R is the value of the second resistor = 6.15 ohm
Therefore:
V = I*R
V = 2.59 * 6.15
V = 15.9285 volts

Hope this helps :)

7 0

3 years ago

HELP ASAP!! i’ll mark you the brainliest!!

Triss [41]

Answer:

yes

Explanation:

6 0

2 years ago

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$