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Zolol [24]
3 years ago
9

A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity

of 20 cms, and if there is no damping, determine the position u of the mass at any time t. Enclose arguments of functions in parentheses. For example, sin(2x).
Physics
1 answer:
nadezda [96]3 years ago
7 0

Answer:

u(t)=\frac{1}{5} sin\ (25t)

Explanation:

Given:

  • mass of the body stretching the spring, m=150\ g
  • extension in spring, \Delta x=1.568\ cm
  • velocity of oscillation, u'(0)=20\ cm.s^{-1}
  • initial displacement position of equilibrium, u(0)=0

<u>According to given:</u>

m.g=k.\Delta x

150\times 980=k\times 1.568

k=93750\ dyne.cm^{-1}

<u>we know frequency:</u>

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{93750}{150} }

\omega=25

Now, for position of mass in oscillation:

u= A.sin\ (\omega.t)+B.cos\ (\omega.t)

u= A.sin\ (25.t)+B.cos\ (25.t)

at t=0;\ u(0)=0\ \Rightarrow A=0

∴u(t)=B.sin\ (25.t)

∵ at t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}

u(t)=\frac{1}{5} sin\ (25t)

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Suppose that the sound level of a conversation is initially at an angry 71 dB and then drops to a soothing 54 dB. Assuming that
Goryan [66]

Answer:

Explanation:

In the decibel scale , intensity of sound changes logarithmically as follows

10log\frac{I}{I_0} = Value in decibel scale , the value of I₀ = 10⁻¹² W /m².

Putting the values

10log\frac{I}{10^{-12}} = 71

log\frac{I}{10^{-12}} = 7.1

\frac{I}{10^{-12}} = 10^{7.1}

I= 10^{-4.9} W/m²

Similarly for 54 dB sound intensity can be given as follows

I = 10⁻¹² x 10^{5.4}

I= 10^{-6.6 } W / m²

For intensity of sound the relation is as follows

I = 2π²υ²A²ρc where υ is frequency , A is amplitude , ρ is density of air and c is velocity of sound .

Putting the given values for 71 dB

I= 10^{-4.9}  = 2π² x 504²xA²x 1.21 x  346

A² = 60.03 x 10⁻¹⁶

A = 7.74 x 10⁻⁸ m

For 54 dB sound

10^{-6.6} = 2π² x 504²xA²x 1.21 x  346

A² = 1.1978 x 10⁻¹⁶

A = 1.1 x 10⁻⁸ m

6 0
3 years ago
on the surface of theearth ,the weight of a boy is 400N but on a mountainpeak his weight is 360N,Calculatethe value of ''g' on t
andreyandreev [35.5K]

The value of g at sea level is 9.81 ms^-2.

The boy's mass is constant wherever he is in the universe but his weight will depend on the strength gravity where he is.

By proportion its value on the mountain peak  is (360 /400) * 9.81

= 0.9 * 9.81 = 8.83  ms^-2  to nearest hundredth,  (answer).

7 0
3 years ago
Using this information...
Pepsi [2]

19.2\:\text{m/s}

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)

\:\:\:\:\:= 19.2\:\text{m/s}

6 0
3 years ago
Help with this please.
gregori [183]
Rutherford's experiment<span> utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that </span>atoms<span> had an inner core which contained most of the mass of an </span>atom<span> and was positively charged.</span>
8 0
3 years ago
According to the law of conservation of matter, what number must be the same on each side of a chemical equation?
olga2289 [7]

Answer:

4. The number of atoms of each element

Explanation:

<em><u>Why is 1 wrong?</u></em>

The number of substances (i.e. number of mols of substances) do not have to be the same. For instance, 2 mol of water reacts with 1 mol of oxygen to form 2 mol of water. Obviously the numer of substances is not conserved. Thus, it isn't a requirement.

<em><u>Why is 2 wrong?</u></em>

The number of chemical symbols do not mean anything. They do not tell us anything about the quantity of matter, they only tell us about the elements that are involved in the reaction.

<em><u>Why is 3 wrong?</u></em>

Once again, the number of chemical formulas are meaningless.

<u><em>Why is 4 correct?</em></u>

The number of atoms are always conserved as atoms cannot be broken or created.

<em><u>Possible confusion...</u></em>

The number of atoms are always conserved but the number of molecules may not always be conserved.

Why? As 2 atoms may be a molecule on the reactant side and 4 atoms may become a molecule on the product size, causing the number of molecules to reduce by half, while the number of atoms remain constant.

Hope that makes sense!

7 0
3 years ago
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