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3 years ago

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When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec

tromagnetic wave of frequency 2Hz. What is the wavelength of this wave?

1 answer:

borishaifa [10]3 years ago

8 0

To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

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

Where,

c = Speed of light (vacuum)

f = frequency

Our values are,

$f = 2Hz$

$c = 3*10^8km/s$

Replacing we have,

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

$\lambda = \frac{3*10^8km/s}{2Hz}$

$\lambda = 1.5*10^8m$

<em>Therefore the wavelength of this wave is $1.5*10^{8}m$ </em>

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3 years ago

Read 2 more answers

An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached

Vera_Pavlovna [14]

Answer:

A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.

B) Magnitude of charges = 1.206 × 10^(-6) C

Explanation:

We are given;

Spring constant;k = 350 N/m

Spring length;L = 0.39 m

Stretched length of spring;x = 0.022 m

A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.

B) Force (F) required to stretch spring is given by the formula;

F = kx

Thus:

F = (350 × 0.022)

F = 7.7 N

Now, if we assume point charges, then the distance (r) between them will be given as:

r = (0.39 + 0.022) = 0.412 m

Coulomb's Law has a formula:

F = k(q1×q2)/r²

where k is coulomb's constant = 8.99 × 10^(9) Nm²/C²

Making q1 × q2 the subject, we have;

(q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))

(q1 × q2) = 14.54 × 10^(-11) C

We are told that both charges are equal, thus; |q1| = |q2|

So;

q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C

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3 years ago

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

What is the top of a wave called?

ohaa [14]

Answer:

The highest part of the wave is called the crest.

Explanation:) hope this helps

The highest part of the wave is called the crest. The lowest part is called the trough. The wave height is the overall vertical change in height between the crest and the trough and distance between two successive crests (or troughs) is the length of the wave or wavelength.

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1 year ago

A 74 kg firefighter slides, from rest, 4.9 m down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that t

In-s [12.5K]

Answer:

Her speed is 9.8 meter per second

Explanation:

Newton's second law states that acceleration (a) is related with force (F) by:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

Here the only force acting on the firefighter is the weight F=mg so (1) is:

$mg=ma$

Solving for a:

$a=g$

Now with the acceleration we can use the Galileo's kinematic equation:

$Vf^{2}=Vo^{2}+2a\varDelta x$ (2)

With Vf the final velocity, Vo the initial velocity and Δx the displacement, because the firefighter stars from rest Vo=0 so (2) is:

$Vf^{2}=2a\varDelta x$

Solving for Vf

$Vf=\sqrt{2g\varDelta x}=\sqrt{2(9.81)(4.9)}$

$Vf=9.8\frac{m}{s}$

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2 years ago