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Aleks [24]
3 years ago
9

The wavelengths of radio waves are much _______________ than the wavelength of microwaves. therefore, radio waves carry much ___

_______ energy than a microwave.
Physics
2 answers:
velikii [3]3 years ago
8 0

Answer:

The wavelengths of radio waves are much longer than the wavelength of microwaves. therefore, radio waves carry much lower energy than a microwave.

Explanation:

The wavelength of radio waves is longer than the wavelength of microwaves.

Using the equation that relates energy and wavelength:

E=\frac{hc}{\lambda}

where h and c are constants (h is Planck's constant and c the constant of the velocity of light)

And \lambda is the wavelength.

What this equation tells us is that the energy is inversely proportional to the wavelength. The larger the wavelength, the lower the energy that the wave carries.

So, since the wavelength of radio waves is much longer, its energy is much lower than the energy of microwaves.

nydimaria [60]3 years ago
7 0
The wavelengths of radio waves are much "Longer" than the wavelength of microwaves therefore, radio waves carry much "Lower" <span>energy than a microwave.

Hope this helps!</span>
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kobusy [5.1K]
In a perfectly ELASTIC collision between two perfectly rigid objects <span>both the momentum and the kinetic energy of the system are conserved.hope it helps</span>
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3 years ago
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An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }

B = \boxed{7.07 *10^{-10} T}

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

\boxed{B = 0 T}

3 0
2 years ago
The equation for the speed of an electromagnetic wave is
igomit [66]

Answer:

c = 1 / √(ε₀*μ₀)

Explanation:

The speed of the electromagnetic wave in free space is given in terms of the permeability and the permittivity of free space by

c = 1 / √(ε₀*μ₀)

where the permeability of free space (μ₀) is a physical constant used often in electromagnetism and ε₀ is the permittivity of free space (a physical constant).

4 0
3 years ago
g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin
DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as mgh=0.5mv^{2} since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then v=\sqrt {2gh} and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s

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IgorLugansk [536]

Answer:

E. downward and constant

Explanation:

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For example, when a ball is thrown up in the air, the ball's velocity is initially upward. Since gravity pulls the object toward the earth with a constant acceleration ggg, the magnitude of velocity decreases as the ball approaches maximum height. At the highest point in its trajectory, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth.

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