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densk [106]
3 years ago
5

It is the obligation of researchers to review and comment on the research of other researchers. True or False.

Physics
2 answers:
Sever21 [200]3 years ago
8 0

It is false that it is the obligation of researchers to review and comment on the research of other researchers. It is not their obligations - they don't have to do it, although they can if they want to and if they are allowed by the author him or herself. However, they are not bound by law or something like that to do this, it's just due to their kindness or genuine interest that they do this.



false

romanna [79]3 years ago
4 0
<span>It is false that it is the obligation of researchers to review and comment on the research of other researchers. It is not their obligations - they don't have to do it, although they can if they want to and if they are allowed by the author him or herself. However, they are not bound by law or something like that to do this, it's just due to their kindness or genuine interest that they do this.</span>
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If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

     v=\sqrt{\frac{T}{(m/l)}}

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

(m/l)=\frac{10}{V^2}

Explanation:

From the question we are told that

Speed of a transverse wave given by

v=\sqrt{\frac{T}{(m/l)}}

Maximum Tension is T=10.0N

Generally making (m/l) subject from the equation mathematically we have

v=\sqrt{\frac{T}{(m/l)}}

v^2=\frac{T}{(m/l)}

(m/l)=\frac{T}{V^2}

(m/l)=\frac{10}{V^2}

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8 0
3 years ago
If two particles have equal kinetic energies, are their momenta necessarily equal? explain.
Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

K.E=\frac{1}{2}mv^{2}

Similarly the momentum is given by m\times v

For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective kinetic energies is given by

K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}

K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}

Similarly For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective momenta are given by

p_{1}=m_{1}\times v_{1}

p_{2}=m_{2}\times v_{2}

Now since it is given that the two kinetic energies are equal thus we have

\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0
3 years ago
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