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hammer [34]
3 years ago
9

Suppose you wanted to change the fundamental frequency of an oscillating string. What are three parameters that you could alter

and how would these alter the oscillation frequency?
Physics
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

Linear density, Length, and tension in the string

Explanation:

The fundamental frequency of an oscillating string is given by:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension in the string

\mu is the linear density of the string, which can also be rewritten as

\mu = \frac{m}{L}

where m is the mass of the string.

Therefore, we can say that in order to change the fundamental frequency of the string, we can change either its lenght, or its tension or its linear density.

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A particle is confined between rigid walls separated by a distance L = 0.167 nm. The particle is in the second excited state (n
ZanzabumX [31]

Answer:

The answer for a classical particle is 0.00595

Explanation:

The equation of the wave function of a particle in a box in the second excited state equals:

ψ(x) = ((2/L)^1/2) * sin((3*pi*x)/L)

The probability is equal to:

P(x)dx = (|ψ(x)|^2)dx = ((2/L)^1/2) * sin((3*pi*x)/L) = (2/L) * sin^2((3*pi*x)/L) dx

for x = 0.166 nm

P(x)dx = (2/0.167) * sin^2((3*pi*0.166)/0.167) * 100 pm = 0.037x10^-3

for x = 0.028 nm

P(x)dx = (2/0.167) * sin^2((3*pi*0.028)/0.167) * 100 pm = 11x10^-3

for x = 0.067 nm

P(x)dx = (2/0.167) * sin^2((3*pi*0.067)/0.167) * 100 pm = 3.99x10^-3

therefore, the classical probability is equal to:

(1/L)dx = (1/0.167)*100 pm = 0.00595

8 0
4 years ago
"a 1,600 kg car is traveling at a speed of 12.5 m/s. what is the kinetic energy of the car?"
xxTIMURxx [149]
I had this question before can you show me the answers choices
6 0
3 years ago
The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track an
pochemuha

Answer:

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

t = 76 s

Find: v

This problem is over-defined.  We only need 3 pieces of information, and we're given 4.  There are several equations we can use.  For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)

v = 3.0 m/s

Or:

Δx = vt − ½ at²

(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

8 0
4 years ago
When does it have kinetic energy
NeTakaya
The answer is motion
5 0
3 years ago
Define escape velocity ​
lianna [129]

Answer:

the lowest velocity which a body must have in order to escape the gravitational attraction of a particular planet or other object.

Explanation:

8 0
3 years ago
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