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Jlenok [28]
3 years ago
13

. When two speakers play different sounds with a constant frequency, beats are heard with a frequency of 5 Hz. If the frequency

of sound played by one of the speakers is 1050 Hz, what is the frequency of sound played by the other speaker
Physics
1 answer:
morpeh [17]3 years ago
3 0

The beat frequency is the concept required to develop this process. The phenomenon is generated when you have two waves but their frequencies do not differ greatly. So the beat frequency will be the difference between those two waves. For this case we have the difference and one of the waves, therefore,

F_b = |F_1-F_2|

Our values are given as,

F_b = 5Hz

F_1 = 1050Hz

Using the formula of the frequency I will have,

F_b = |F_1-F_2|

Replacing we have,

5 = |1050-F_2|

F_2 = 1050 \pm 5

F_2 = 1045,1055 Hz

The possible values are two: 1045Hz and 1055Hz

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Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

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Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is:   w_{c} =m_{c} g   and maximum tension is T=2.50 m_{c} g=1.41*10^4N

total mass and weight is :

M =m_{c}+ m_{b} =740kg+550kg=1290 kg

w_{M} =1.2650*10^4N

∑F_{y} =ma_{y}

T-M_{g} =Ma_{y}

a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg

=0.645m/s^2

B)

maximum acceleration

a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0

using y-y_{0} =v_{oy} t+1/2(a_{y} )t^2

to solve for t

t=\sqrt{2(y-y_{0} )/a_{y} }

t=\sqrt{2(119m)/0.645m/s^2} =19.20s

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A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur
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Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block  from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block  from the hole when the block is revolved=9\times 10^{-2}m

a.We have to find the tension in the cord in the original situation when the block has speed =v_0=0.63 m/s

T=\frac{mv^2}{r}

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T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N

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c.Work don=Final K.E-Initial K.E

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Answer:

Net force on the block is 32 N.

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Explanation:

Let the acceleration of the object be a m/s².

Given:

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Frictional force on the block is, f=8\ N

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

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