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Arada [10]
3 years ago
9

A 5.36 kg object falls freely (ignore air resistance), after being dropped from rest. Determine the initial kinetic energy (in J

), the final kinetic energy (in J), and the change in kinetic energy (in J) for the following.
(a) first meter of fallinitial kinetic energy? Jfinal kinetic energy? Jchange in kinetic energy? J(b) second meter of fallinitial kinetic energy? Jfinal kinetic energy? Jchange in kinetic energy
Physics
1 answer:
matrenka [14]3 years ago
6 0

Answer: 52.53

Explanation:

m*g = 5.36k * 9.8N/kg = 52.53 N. = Wt.

of object.

a. KE = 0 J. = Initial KE

V^2 = Vo^2 + 2g*h

V^2 = 0 + 19.6*1 = 19.6

V = 4.43 m/s.

KE = 0.5m*V^2 = 2.68*4.43^2 = 52.59 J.

b. V^2 = Vo^2 + 2g*h

V^2 = 4.43^2 + 19.6*1 = 39.22

V = 6.26 m/s.

KEo = 2.68*4.43^2 = 52.59 J.

KE = 2.68*6.26^2 = 105 J.

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In the writing of ionic chemical formulas the value of each ion's charge is crossed over in the crossover rule.

Rules for naming Ionic compounds

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A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

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Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

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Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

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Answer:

Answer to the question:

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