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WARRIOR [948]
2 years ago
5

A squeeze bottle squeezes when pressed. It regains its shape when pressed .It regains its shape when the pressure from your hand

is withdrawn. What may happen if the squeeze bottle is pressed to take the sauce out and then immediately corked tightly? Will it regain its shape? If not, Why?
Physics
1 answer:
Leokris [45]2 years ago
5 0

Answer:

The squeeze will not regain its shape

Explanation:

The squeeze bottle will not regain its shape.

This is because the atmospheric pressure compresses the squeeze bottle. Since the pressure in the squeeze bottle is now not equal to the atmospheric pressure since it has been corked tightly, its internal pressure cannot balance out the atmospheric pressure and thus cancel its effect.

So, the squeeze bottle does not regain its shape due to this imbalance of pressure.

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A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the
Sedbober [7]
Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
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3 0
2 years ago
Read 2 more answers
Q6. The 600 N force applied to the bracket at As to be replaced by the two forces, Fa in the a-a direction and Fo in the b-b dir
jeka94

answer:

The horizontal component of the 600 N force is,

Fx = 600 cos(30) 519.61 N

The vertical component of the 600 N force is,

Fy = 600 sin(30) = 300 N

hope it's help!

6 0
1 year ago
The frequency of the middle d note on a piano is 293.66 hz. what is the wavelength of this note in centimeters? the speed of sou
Natali5045456 [20]
Wavelength = speed of sound / frequency

= (343.06 m/s) / (261.63 Hz)

= 1.31 m (or 131 cm)
4 0
2 years ago
You are exploring a planet and drop a small rock from the edge of a cliff. In coordinates where the +y direction is downward and
Lelu [443]

Answer:

value of the acceleration of gravity on the planet is 5.00 m/s²

Explanation:

The problem is similar to a free fall exercise, with another gravity value, the expression they give us is the following:

       y-yo = ½ gₐ t²       (1)

They tell us that they make a squared time graph with the variation of the distance, it is appropriate to clarify this in a method to linearize a curve, which is plotted the nonlinear axis to the power that is raised, specifically, the linearization of a curve The square is plotted against the other variable.

  Let's continue our analysis, as we have a linear equation, write the equation of the line.

     

        y1 = m x1 + b       (2)

where  “y1” the dependent variable, “x1” the independent variable, “m” the slope and “b” the short point

In this case as the stone is released its initial velocity is zero which implies that b = 0,

We plot on the “y” axis the time squared “t²” and on the horizontal axis we place “y-yo”.  To better see the relationship we rewrite equation 1 with this form

        t² = 2 /gₐ  (y-yo)

 

With the two expressions written in the same way, let's relate the terms one by one

        y1 = t²

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We substitute and calculate

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This is the value of the acceleration of gravity on the planet, note that the decimals are to keep the figures significant

6 0
2 years ago
When a 75.0-kg man slowly adds his weight to a vertical spring attached to the ceiling, he reaches equilibrium when the spring i
Sergio039 [100]

Answer:

1)k=11.319kN/m

2)displacement=13.02cm

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Explanation:

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thus we have at equilibrium

kx=mg\\\\k=\frac{mg}{x}

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k=\frac{75\times 9.81}{0.065}\\\\k=11.319kN/m

2)

When we add another identical spring we get an equivalent spring with spring constant as  

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Thus at equilibrium we have

x_{2}k_{eq}=mg\\\\x_{2}=\frac{mg}{k_{eq}}\\\\x_{2}=\frac{75\times 9.81}{5.65}\times 10^{-3}=13.02cm

3) Equivalent spring constant will be as calculated earlier k_{eq}=5.65kN/m

3 0
2 years ago
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