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2 years ago

Describe what a high and low frequency electromagnetic wave look like:

1 answer:

4 0

A high electromagnetic wave has short, very fast, frequent waves.

a low electromagnetic wave has long, very slow, infrequent waves.

hope this helps! pls mark brainliest!

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The air in living room has a mass of 72 kg and a specific heat of 1010 J/(kg°C). What is the approximate change in thermal energ

ICE Princess25 [194]

Answer:

Explanation:

3 0

2 years ago

Using the work energy theorem. What is the velocity of a 850kg car after starting at rest when 13,000J of work is done to it.

Misha Larkins [42]

When the car starts moving it acquires kinetic energy.

For this problem the formula of kinetic energy is:

$\frac{1}{2}mv^2$

Where m is the mass of the car = 850kg, and v is the speed of the car.

If we consider insignificant the energy lost by friction with soil and air we can propose the following equation:

Applied energy = Kinetic energy acquired by the car.

$13000\ J = \frac{1}{2}mv^2\\\\13000\ J = \frac{1}{2}(850)v^2\\\\\frac{13000(2)}{850} = v^2\\\\v = \sqrt{\frac{13000(2)}{850}}\\\\v = 5.531 m/s$

The velocity is 5.531 m/s

3 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0

3 years ago

The period of a carrier wave is T=0.01 seconds. Determine the frequency and wavelength of the carrier wave. a. f=10 Hz, λ=3E8 me

makvit [3.9K]

Explanation:

It is given that,

The period of the carrier wave, T = 0.01 s

Let f and $\lambda$ are frequency and the wavelength of the wave respectively. The relationship between the time period and the frequency is given by :

$f=\dfrac{1}{T}$

$f=\dfrac{1}{0.01}$

f = 100 Hz

The wavelength of a wave is given by :

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

$\lambda=\dfrac{3\times 10^8}{100}$

$\lambda=3\times 10^6\ m$

So, the frequency and wavelength of the carrier wave are 100 Hz and $3\times 10^6\ m$ respectively. Hence, the correct option is (c).

6 0

2 years ago

A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an

Nadya [2.5K]

Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm

Answer: 4.5 cm

7 0

3 years ago

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