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3 years ago

A unit for measuring frequency is the A. hertz. B. watt. C. ampere. D. ohm.

2 answers:

5 0

One hertz means that an event repeats one per second so the unit for measuring frequency is the hertz and the answer is A.hertz :)))
i hope this be helpful

baherus [9]3 years ago

3 0

The answer is A. Hertz. Frequency is the number of occurences of an event per unit of time. Hertz are commonly used to measure radio waves, which are frequencies, so the answer must be hertz.

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A polar bear walks 27 km south then 17 km east. What is the polar bears final displacement?

Brilliant_brown [7]

It would be 44km to the east

6 0

3 years ago

A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp

Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

$y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}$

1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

2. Finding time t at position y = 1 cm:

$0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}$

3. Find velocity v at time t from equation 2:

$v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))$

5 0

3 years ago

Read 2 more answers

Any example of acceleration related to human body??? <br> answer quickly plz :)

Hitman42 [59]

Answer:

Acceleration stress, physiological changes that occur in the human body in motion as a result of rapid increase of speed. ... A force of 3 g, for example, is equivalent to an acceleration three times that of a body falling near Earth.

4 0

3 years ago

Two large parallel conducting plates are 8.0 cm apart and carry equal but opposite charges on their facing surfaces. The magnitu

Pepsi [2]

Answer:

18.1 V

Explanation:

The electric field between two parallel plates is given by the equation:

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma$ is the charge surface density

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

For the plates in this problem,

$\sigma = 2.0 nC/m^2 = 2.0\cdot 10^{-9} C/m^2$

So, the magnitude of the electric field is

$E=\frac{2.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=226.0 V/m$

Now we can find the potential difference between the plates, which is given by

$\Delta V = E d$

where

d = 8.0 cm = 0.08 m is the separation between the plates

Substituting,

$\Delta V=(226.0)(0.08)=18.1 V$

8 0

3 years ago

Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of

julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b) E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro, therefore we can consider the charge to be concentrated in the center

E₁ = k q₁ / d²

shell 2

the point is on the outside,d>ro

E₂ = k q₂ / d²

the total camp is

E_total = -E₁ + E₂

E_total = k ( $\frac{-q_1 + q_2}{d^2}$ )

E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

E₂ = 0

E_total = E₁

E_total = k q₁ / d²

E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point is inside the shell d< ro

as there are no charges inside

E₁ = 0

shell 2

point is inside the radius of the shell d <ro

E₂ = 0

the total field is

E_total = 0

3 0

3 years ago