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Llana [10]
2 years ago
11

A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is trans- formed i

nto elastic potential energy of the bending board, if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the fore. Ultimately the board breaks at an applied force of 800 N and a deflection of 1.2 cm.
a. To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand.
b. If the hand is moving this fast and comes to rest in a dis- tance of 1.2 cm, what is the average force on the hand?
Physics
1 answer:
ra1l [238]2 years ago
3 0

Answer:

a) v = 4.4 m/s

b) F = 400 N

Explanation:

a) ½kx² = ½mv²

v = √(kx²/m)

F = kx

v = √(Fx/m)

v = √(800(0.012) / 0.5) = √19.2 = 4.3817...

b) Fd = ½mv²

F = mv²/2d

F = 0.5(19.2) / (2(0.012) = 400 N

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To develop this problem, it is necessary to apply the concepts related to Beat

The Beat is an acoustic phenomenon that is generated by two sine waves interfering with slightly different frequencies. The beat frequency is equal to the difference in the frequencies of the two original waves:

f_{bat} = |(f_1 \pm f_B)|

Our values are given as

f_1 = 440Hz

f_B = 2.1Hz

For the particular case we have two possible frequencies:

1) f_{bat} = |(f_1 - f_B)|

f_{bat} = |(440 - 2.1)|

f_{bat} = 437.9Hz

2) f_{bat} = |(f_B + f_1)|

f_{bat} = |(440 +2.1)|

f_{bat} = 442.1Hz

Therefore the two possibles frequencies of the other players note are 437.9Hz and 442.1Hz

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Based on its position in the periodic table, you can predict how many electrons it has, how many valence electrons, how many levels of electrons, and its atomic number and mass.
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3 years ago
The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are
Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

  • Sin(34°) = <em>y</em> / 10.8

Then we <u>solve for </u><u><em>y</em></u>:

  • 0.559 = <em>y</em> / 10.8
  • <em>y </em>= 6.0

And for the horizontal component, we use the formula:

  • Cos(34°) = <em>x</em> / 10.8

Then we <u>solve for </u><u><em>x</em></u><u>:</u>

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2 years ago
This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a
Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

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3 years ago
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