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Alik [6]
3 years ago
9

How much pressure is exerted by an 100n man with a shoe area of O.05cm square​

Physics
1 answer:
DIA [1.3K]3 years ago
5 0

Answer:

Pressure = 20 MPa

Explanation:

Given:

Force acting on the shoe is, F=100\ N

Area of shoe on which the force acts is, A= 0.05\ cm^2

Now, first we convert the area into its standard unit of m².

We have the conversion factor as:

1 cm² = 10^{-4}\ m^2

Therefore, the area of shoe in square meters is given as:

A=0.05\times 10^{-4}\ m^2\\A=5\times 10^{-6}\ m^2

Now, pressure on the shoe is given as:

P=\frac{Force}{Area}\\P=\frac{F}{A}

Plug in 100 N for 'F', 5\times 10^{-6} for 'A' and solve for 'P'. This gives,

P=\frac{100\ N}{5\times 10^{-6}\ m^2}\\P=20\times 10^{6}\ N/m^2

Now, we know that,

10^{6}\ N/m^2=1\ MPa\\\therefore 20\times 10^{6}\ N/m^2=20\times 1\ MPa=20\ MPa

Therefore, the pressure acting on the shoe is 20 MPa.

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The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s  first law, the net force on gymnast,

F_{net} =F-W=ma

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast (9\times g acceleration due to gravity)  

Therefore,

F= ma+W OR F=ma+mg=m(g+a)

Given m = 30 kg anda=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}

Substituting these values in above formula and calculate the force exerted by the gymnast,  

F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )

F=3.537\times10^{3}N

6 0
3 years ago
A physical count of merchandise inventory on november 30
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3 years ago
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

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

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

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

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

6 0
3 years ago
Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them
denpristay [2]

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

3 0
3 years ago
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jeka94

Answer:

This question will be answered based on general photosynthetic understanding. The answer is:

The production of oxygen would increase

Explanation:

The characteristics of most plant forms is their ability to photosynthesize i.e. use solar energy (from sunlight) to make food (chemical energy). The product of this photosynthetic process is OXYGEN gas, which is released as a waste product via the stomata on their leaves. Note that, photosynthesis cannot occur without LIGHT as it provides the energy needed for the process.

Hence, in the duckweed plant like every other photosynthetic plant, the increase in the intensity and duration of exposure to light means the rate at which photosynthesis occurs will be increased. An increased photosynthetic rate means the synthesis of the products will also be increased i.e. glucose and OXYGEN.

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