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ExtremeBDS [4]
2 years ago
7

Being timed help!!!

Physics
1 answer:
Virty [35]2 years ago
8 0

The wavelength of the 5th harmonic will be 145 cm. None of the given options are correct.

<h3>What is wavelength?</h3>

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

The length of the string corresponds to one-half of a wavelength for the first harmonic.

\rm \lambda_1= \frac{1}{2} L \\\\ \lambda_5=5 \times  \frac{1}{2} L  \\\\\ \lambda_5=5 \times  \frac{1}{2} \times 58 \ cm  \\\\\   \lambda_5 = 145 \ cm

Hence the wavelength of the 5th harmonic will be 145 cm.

To learn more about the wavelength refer to the link;

brainly.com/question/13533093

#SPJ1

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Frequency is the ____________ that move past a point during a specific amount of time. Frequency is measured in ____________ , a
yulyashka [42]

Answer:

Frequency is <u>the number of waves</u> that move past a point during a specific amount of time. Frequency is measured in <u>Hertz</u>, and is classified as high, medium, or low. Frequency is interpreted as the <u>pitch</u> of a sound. Intensity refers to the <u>loudness</u> of a sound and is measured in <u>decibels</u>. Louder sounds <u>increase</u> the rate of nerve signals relayed to the brain.

Explanation:

6 0
4 years ago
A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl
Sati [7]

Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

6 0
4 years ago
Please help me with this
aev [14]

Answer:

the answer is UV Radiation

6 0
3 years ago
A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i
Hoochie [10]

The acceleration of the box is approximately 1 m/s^2

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

\sum F = F_a - F_f

where

F_a = 20 N is the applied forward force

F_f=9.0 N is the friction force

Combining the two equations,

F_a-F_f=ma

And solving for the acceleration,

a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

8 0
3 years ago
A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang
valina [46]

Answer:

work done is -2.8  × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta

W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta

W = -mgl  \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta

W = -mgl[ -cosθ ]^{0.047}_{0.045 }

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8  × 10⁻⁶ J

Therefore, work done is -2.8  × 10⁻⁶ J

6 0
3 years ago
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