PHYSICS COLLEGE

If a microwave oven produces electromagnetic waves with a frequency of 2.30 ghz, what is their wavelength?

Answers

Answer 1

Answer:

Answer: wavelength is $<strong> 1.30 * 10^8 nm.</strong>$

The frequency of the microwave is, f = 2.30 GHz.

To Find frequency use the formula:

$c=f$ λ

Where, c is the speed of electromagnetic wave or light. f is the frequency, and λ is the wavelength of light.

Rearranging, $\lambda = (c)/(f)$

Plug in the values,

$\lambdam = (3 * 10^8 m/s)/(2.30 GHz(10^9 Hz)/(1 GHz))=0.130 m(10^9 nm)/(1 m) = 1.30 * 10^8 nm.$

Answer 2

Answer: $\lambda = (c)/(\nu) = (3 \cdot 10^8)/(2.3 \cdot 10^9) = (3)/(23) \approx 0.130435 \approx 0.13 \ m.$

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The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally from west to east. the vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 3.00 10-5 t. (a) in what direction are the electrons deflected by this field component? due north due south due east due west (b) what is the magnitude of the acceleration of an electron in part (a)? m/s2

Answers

(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
$K= (1)/(2)mv^2$
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
$v= \sqrt{ (2K)/(m) }= \sqrt{ (2 \cdot 2.20 \cdot 10^(-15) J)/(9.1 \cdot 10^(-31) kg) }=6.95 \cdot 10^7 m/s$

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
$qvB = m (v^2)/(r)$
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
$r= (mv)/(qB)= ((9.1 \cdot 10^(-31) kg)(6.95 \cdot 10^7 m/s))/((1.6 \cdot 10^(-19) C)(3.00 \cdot 10^(-5) T))=13.18 m$

And finally we can calculate the centripetal acceleration, given by:
$a_c = (v^2)/(r)= ((6.95 \cdot 10^7 m/s)^2)/(13.18 m)=3.66 \cdot 10^(14) m/s^2$

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Answers

To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_(1,2)$ = Mass of each object

$v_(1,2)$ = Initial Velocity of each object

$v_f$ = Final velocity

Replacing we have that,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

$1850*13.8+3100*0 = (1850+3100)v_f$

$v_f = 5.1575m/s$

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

$v_f^2-v_i^2 = 2ax$

Since there is no initial speed, then

$v_f^2 = 2ax$

$5.1575^2 = 2a (1.91)$

$a = 6.9633m/s^2$

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

$F_f = F_a \n\mu N = m*a \n\mu = (ma)/(N)\n\mu = (ma)/(mg)\n\mu = (a)/(g)\n\mu = (6.9633)/(9.8)\n\mu = 0.7105$

Therefore the Kinetic friction coefficient is 0.7105

A beam of light, which is traveling in air, is reflected by a glass surface. Does the reflected beam experience a phase change, and if so, by how much is the phase of the beam changed?

Answers

The reflected beam experienced a phase change of about 180°.

What is reflection in the glass surface?

According to Snell's law, the light that incident on the glass surface will be reflected and transmitted at an angle equals to the angle of incidence.

By the observation of refractive index of the glass for the normal incidence only 4% of the light is transmitted or reflected.

The light passing through glass is not only reflected on the front surface, but also on the back. For several times the light will gets reflected back and forth. So, the total reflectance through a glass window can be calculated as

2·R / (1+R).

Thus, A light wave travelling in air is reflected by a glass barrier will undergo a phase change of 180°, while light travelling in glass will not undergo a phase change if it is reflected by a boundary with air.

Learn more about reflection,

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Answer:

180 degree phase change

Explanation:

A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s . What is her acceleration on the rough ice?

Answers

A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. Her acceleration on the rough ice is -3.65 m/s².

v² - u² = 2 a ∆x, where u and v are initial and final velocities, respectively; a is acceleration.

and ∆x is the distance traveled (because the skater moves in only one direction).

Thus, (5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s².

Thus, A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. Her acceleration on the rough ice is -3.65 m/s².

Learn more about Acceleration, refer to the link:

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Recall that

v² - u² = 2 a ∆x

where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled (because the skater moves in only one direction).

So we have

(5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s²

A long wire carries a current toward the south in a magnetic field that is directed vertically upward. What is the direction of the magnetic force on the wire?

Answers

Answer:

the answer is it is going north

Explanation:

because its the opposite

Final answer:

The magnetic force on a wire carrying current towards the south under a magnetic field directed vertically upwards will point towards the East. In order to determine this, use the right-hand rule.

Explanation:

The direction of the magnetic force on a current-carrying wire under a magnetic field can be deduced using the right-hand rule. In this case, with the current flowing towards the south and the magnetic field directed vertically upward, you would point your right thumb in the direction of the current (southwards) and curl your fingers in the direction of the magnetic field (upwards). The palm of your hand will then face toward the direction of the force. In this case, the force would be pointing toward the East.

The right-hand rule is a vital principle in the study of electromagnetism as it aids in identifying the direction of various quantities in magnetic fields. The magnetic force on a current-carrying wire represents the phenomenon underlying the working of many electric motors.