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lilavasa [31]
2 years ago
9

A string has a mass of 0.8Kg. The length of the string is 2m. The tension in the string is 5N. The string is stretched and fixed

at both ends such that when struck it forms the second harmonic. Calculate the frequency of the wave which results on the string.
Physics
1 answer:
viktelen [127]2 years ago
3 0

Answer:

1.25Hz

Explanation:

For waves on a string, the second harmonic is obtained from;

2fo = 1/l √T/M

Where;

l = length of the string

M= mass in kilograms

T = tension in the string

2fo = 1/2√5/0.8

2f0 = 1.25Hz

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After robbing a bank, a criminal tries to escape from the police by driving at a constant speed of 55 m/s (about 125 mph). A pol
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Answer:

18.03 s

Explanation:

We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.

x=vt\\x=x_{0}+v_{0}t+\frac{a}{2}t^2

The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:

x=(55m/s)t\\x=\frac{6.1m/s^2}{2}t^2=(3.05m/s^2)t^2

Equalizing these two equations and solving for t:

(55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=\frac{55m/s}{3.05m/s^2}=18.03 s

6 0
3 years ago
Two long, straight wires, one above the other, are seperated by a distance 2a2a and are parallel to the x−axisx−axis. Let the +y
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Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
3 0
3 years ago
Two long, straight parallel wires are placed 38 cm apart, one above the other. The top and bottom wires are carrying currents 4.
ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

Given;

distance between the two parallel wire, d = 38 cm = 0.38 m

current in the first wire, I₁ = 4.0 kA

current in the second wire, I₂ = 8.0 kA

Force per unit length, between two parallel wires is given as;

\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }

where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

Substitute the given values in the above equation and calculate the force per unit length

\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m

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Hari planted to go abroad. change into negative​
marishachu [46]

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