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3 years ago

What is the frequency of a microwave that has a wavelength of 0.050 m?

Physics

2 answers:

Elanso [62]3 years ago

5 0

Answer:

$6x10^9Hz$

Explanation:

we use a formula that relates the frequency and the wavelength:

$f=\frac{c}{\lambda}$

where $f$ is the frequency of the wave, $c$ is the speed of light $c=3x10^8m/s$ , and $\lambda$ is the wavelength.

We know that the wavelength is: $\lambda=0.050m$ , so substituting the known values in the equation for the frequency we get:

$f=\frac{3x10^8m/s}{0.05m} \\f=6x10^9s^{-1}=6x10^9Hz$

The frequency is:

$6x10^9Hz$

Fudgin [204]3 years ago

4 0

Frequency Wavelength 1/4 Wavelength 1/20 Wavelength 1/100 Wavelength
600 MHz 0.5 meters 12.5 cm 2.5 cm 5.0 mm

if that helps at all

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Which statements accurately describe Dmitri Mendeleev’s contributions to the development of the periodic table? Check all that a

Levart [38]

He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements

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3 years ago

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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

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3 years ago

El dormitorio de Pablo es rectangular, y sus lados miden 3 y 4 metros. Ha decidido dividirlo en dos partes triangulares con una

Paraphin [41]

Answer:

c = 5 m

Explanation:

this exercise you want to divide the rectangular room into two triangular rooms

the area of triangles is

A = ½ base height

A = ½ 4 3

A = 6 m²

the length of the curtain can be found using the Pythagorean theorem

c² = b² + a²

c = √ (4² + 3²)

c = 5 m

this is the length of the curtain

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3 years ago

A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the

barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:

$KE=\frac{kQq}{r}\ \ \ \ \ \ \ ...i$

$k$ is Coulomb's constant.

#The electric field, $E$ at radius r is expressed as:

$E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii$

From i and ii, we have:

$KE=Eqr$

$r=(KE)/Eq$

#Substitute actual values in our equation:

$r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m$

Hence, the distance between the charge and the source of the electric field is 2.2 meters

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3 years ago

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a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

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4 years ago