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Semenov [28]
3 years ago
12

An oscillator creates periodic waves on two strings made ofthe same material. The tension is the same in both strings.If the str

ings have different thicknesses,which of the following parameters, if any, will be different in thetwo strings?Check all that apply.a. wave frequencyb. wave speedc. wavelengthd. none of the above
Physics
1 answer:
Novay_Z [31]3 years ago
7 0

Answer:

c. wavelength

Explanation:

The speed of the wave on the string is given by

v= \sqrt{\frac{T}{\mu} }

Here, \mu is the mass per unit length and T is the tension in the string.

For the different thickness, the mass per unit length is different. Therefore, the wave speed is different in the two strings.

The frequency of the oscillations depends upon the oscillator. So, the frequency is same for the two strings by using same oscillator.

The frequency and speed relation is,

f= γλ

λ= f/γ

Since frequency is constant, the wavelength of the waves different as the speeds are different.

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An electron of mass 9.11×10−31 kgkg leaves one end of a TV picture tube with zero initial speed and travels in a straight line t
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Explanation:

We will use the equations of constant acceleration to find out a_{x} and time t.

As we know that the initial speed is zero. So

(a)  

v_{0x} = 0

x - x_{o} = 1.25×10^{-2}m

v_{x} = 3.3×10^{6}m/s

v^{2} _{x} =  v^{2} _{x_{o} } + 2a_{x} (x - x_{o} )

a_{x} = \frac{v^{2} _{x} - v^{2} _{ox} }{2(x - x_{o}) }

   = \frac{(3.3 * 10^{6})^{2}  - 0 }{2(1.25 * 10^{-2}) }

   = 4.356×10^{14} m/s²

(b)

v_{x} = v_{ox} + a_{x}t

t = v_{x} - vo_{x}/a_{x}

t = \frac{3.00 * 10^{6} }{4.356*10^{14} } = 6.8870×10^{-9}s

(c)

ΣF_{x} = ma_{x}

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3 years ago
A stone is dropped from the upper observation deck of a tower, 250 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dista
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(a) y(t)=250 - 4.9 t^2

For an object in free-fall, the vertical position at time t is given by:

y(t) = h + ut - \frac{1}{2}gt^2

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

y(t) = h - \frac{1}{2}gt^2=0

So for the stone in the problem, we have

250 - 4.9 t^2 = 0

Solving for t, we find:

t=\sqrt{\frac{250}{4.9}}=7.14 s

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

v(t) = u - gt

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s

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