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Rashid [163]
3 years ago
6

What is the mathematical relationship between wavelength and velocity? Inverse, horizontal, linear or quadratic? I NEED MAJOR HE

LP PLZ HELP ME THANK U
Physics
2 answers:
Tju [1.3M]3 years ago
5 0

Answer: Wavelength is the measure of the length of a complete wave cycle. The velocity of a wave is the distance traveled by a point on the wave. In general, for any wave the relation between Velocity and Wavelength is proportionate. It is expressed through the wave velocity formula.

Explanation: For any given wave, the product of wavelength and frequency gives the velocity. It is mathematically given by wave velocity formula written as-

V=f×λ

Where,

V is the velocity of the wave measure using m/s.

f is the frequency of the wave measured using Hz.

λ is the wavelength of the wave measured using m. Velocity and Wavelength Relation

Amplitude, Frequency, wavelength, and velocity are the characteristic of a wave. For a constant frequency, the wavelength is directly proportional to velocity.

Given by:

V∝λ

Example:

For a constant frequency, If the wavelength is doubled. The velocity of the wave will also double.

For a constant frequency, If the wavelength is made four times. The velocity of the wave will also be increased by four times.

Hope you understood the relation between wavelength and velocity of a wave. I truely hope this helps you out tho! Goodluck!

cupoosta [38]3 years ago
4 0
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Blue bounces off the paper
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After a nucleus undergoes radioactive decay, its new mass number is:
Ivanshal [37]
Radioactive "decay" means particles and stuff shoot OUT of a nucleus.
After that happens, there's less stuff in the nucleus than there was before.
So the new mass number is always less than the original mass number.
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4 years ago
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Three children are riding on the edge of a merry-go-round that is 182 kg, has a 1.60 m radius, and is spinning at 15.3 rpm. The
Korolek [52]

Answer:

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

Explanation:

The merry-go-round can be represented by a solid disk, whereas the three children can be considered as particles. Since there is no external force acting on the system, we can apply the principle of angular momentum conservation:

\left(\frac{1}{2}\cdot M + m_{1}+m_{2} + m_{3} \right)\cdot R^{2}\cdot \dot n_{o} = \left(\frac{1}{2}\cdot M + m_{1} + m_{3})\cdot R^{2}\cdot \dot n_{f} (1)

Where:

M - Mass of the merry-go-round, in kilograms.

m_{1}, m_{2}, m_{3} - Masses of the three children, in kilograms.

R - Radius of the merry-go-round/Distance of the children with respect to the center of the merry-go-round, in meters.

\dot n_{o}, \dot n_{f} - Initial and final angular speed, in revolutions per minute.

If we know that M = 182\,kg, m_{1} = 17.4\,kg, m_{2} = 28.5\,kg, m_{3} = 32.8\,kg, R = 1.60\,m and \dot n_{o} = 15.3\,\frac{rev}{min}, then the final angular speed of the system is:

\dot n_{f} = \dot n_{o}\cdot \left(\frac{\frac{1}{2}\cdot M + m_{1} + m_{2} + m_{3} }{\frac{1}{2}\cdot M + m_{1} + m_{3} } \right)

\dot n_{f} = \left(15.3\,\frac{rev}{min} \right)\cdot \left[\frac{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg +28.5\,kg + 32.8\,kg }{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg + 32.8\,kg } \right]

\dot n_{f} = 18.388\,\frac{rev}{min}

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

7 0
3 years ago
A drag racer crosses the finish line doing 212 mi/h and promptly deploys her drag chute. (A.) what force must the drag chute exe
Svet_ta [14]

initial speed of the racer is given as

v_i = 212 mi/h

v_i = 212*\frac{1609}{3600} = 94.75 m/s

after applied force the final speed is given as

v_f = 45 mi/h

v_f = 45 * \frac{1609}{3600} = 20.11 m/s

now during this speed change the racer will cover total distance 185 m

so here we will use kinematics

v_f^2 - v_i^2 = 2 a d

20.11^2 - 94.75^2 = 2*a*185

a = -23.2 m/s^2

now the force that chute will exert on the racer will be given as

F = ma

F = 898* 23.2

F = 2.1* 10^4 N

B) here following is the strategy for solving it

1. first we used kinematics to find the acceleration of the car

2. then we used Newton's II law (F = ma) to find the force

4 0
3 years ago
Help me with this problem please
zaharov [31]

Answer:

Total moment of inertia when arms are extended: 1.613 kg\,m^2

Explanation:

This second part of the problem could be a pretty complex one, but if they expect you to do a simple calculation, which is what I imagine, the idea is just adding another moment of inertia to the first one due to the arms extended laterally and use the moment of inertia for such as depicted in the image I am attaching.

In that image:

L is the length from one end to the other of the extended arms (each 0.75m from the center of the body) which gives 1.5 meters.

m is the mass of both arms. That is: twice 5% of the mass of the person: which mathematically can be written as: 2 * 0.05 * 56.5 = 5.65 kg

Therefore this moment of inertia to be added can be obtained using the formula shown in the image:

I_z=\frac{1}{12} \,m\,L^2\\\\I_z=\frac{5.65\,*\,1.5^2}{12} \\I_z=1.05937\,kg\,m^2

Now, one needs to add this to the previous moment that you calculated, resulting in:

0.554 + 1.059 = 1.613 kg\,m^2

5 0
3 years ago
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