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Vesna [10]
3 years ago
14

When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin

g is also clamped at its midpoint, the lowest four resonant frequencies are __________.
Physics
1 answer:
mario62 [17]3 years ago
4 0

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

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What kind of mirrors are typically used for security purposes?
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Answer:

CONVEX

Explanation:

BECOZ IT HELPS TO GET A WIDER VIEW

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4 years ago
A coin is dropped from a height of 188 meters. Neglecting air resistance, how much time (in seconds) was the coin in the air?
Marina CMI [18]

Answer:

t=6.19s

Explanation:

With a initial velocity is zero, kinematics equation is:

y=1/2*g*t^2

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3 years ago
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4 years ago
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What is the change in internal energy if 30 J of thermal energy is released
alexira [117]

Answer:

ΔU = -70 J

Explanation:

ΔU = Q − W

where ΔU is the change in internal energy,

Q is the heat absorbed by the system,

and W is the work done by the system (on the surroundings).

30 J of thermal energy is released, so Q = -30 J.

40 J of work is done by the system, so W = 40 J.

Therefore, the change in internal energy is:

ΔU = -30 J − 40 J

ΔU = -70 J

6 0
3 years ago
Force f⃗ =−10j^n is exerted on a particle at r⃗ =(7i^+5j^)m. part a what is the torque on the particle about the origin?
cluponka [151]

Answer:

Torque, \tau=0i+0j-70k

Explanation:

It is given that,

Force acting on the particle, F=-10j\ N

Position of the particle, r=(7i+5j)\ m

We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :

\tau=r\times F

\tau=(7i+5j)\times (-10j)

The cross product of vectors is given by :

\tau=\begin{pmatrix}0&0&-70\end{pmatrix}

or

\tau=0i+0j-70k

So, the torque on the particle about the origin 0i+0j-70k. Hence, this is the required solution.

6 0
3 years ago
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