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ladessa [460]
3 years ago
14

a 645 N person is wearing stiletto shoes. If she lifts her left leg, and rocks back onto the heel under her right leg, the press

ure the heel exerts on the floor is 9000000 PA. what is the diameter of the circular base of the heel in centimeters?
Physics
1 answer:
fgiga [73]3 years ago
7 0

Answer:

0.96cm

Explanation:

Given parameters:

Weight of person = 645N

Pressure of the heel = 9000000PA

Unknown

Diameter of the circular base = ?

Solution

Pressure is the force per unit area of a body.

  Pressure = \frac{Force}{Area}

To solve this problem, find the unknown area of the circle;

             Pressure x Area = Force

         Area     = \frac{Force}{Pressure}

           Area  = \frac{645}{9000000} = 7.17 x 10⁻⁵m²

Note: 1Pa = 1Nm⁻²

Now;

  Area of a circle = π r²

         r = \frac{d}{2}

    Area =  π x \frac{d^{2} }{4}

d is the diameter

                    d²  = 4 x Area  x (1/π)

                d² = 4 x 7.17 x 10⁻⁵ x  (1/3.14) = 9.13 x 10⁻⁵

              d = 0.0096m

To cm;

                100cm = 1m

   0.0096m; 0.0096 x 100=  0.96cm

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A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
Alex

Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

Explanation:

pressure on the submarine P_{sea} = 62 MPa = 62 x 10^6 Pa

we also know that P_{sea} = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

P_{sea} = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine P_{g} = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure P_{sea}, we have

62 x 10^6 = 10094.49h

depth h = <em>6141.96 m</em>

7 0
3 years ago
A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050
Westkost [7]

Answer:

The current in wire resistance 2Ω

a). 8696 A

b). fraction power 15.1% a 115kV

Explanation:

Resistance

R=0.05Ω/Km*40km

R=2Ω

P=1000 MW

a).

P=V*I\\I=\frac{P}{V}=\frac{1000x10^{6}W}{115x10^{3}k }  =8696.65A

Using law ohm

b).

V=I*R\\I=\frac{V}{R}

P=I*I*R\\P=I^{2} *R\\P=8696.65^{2}*2\\P=151.228 x10^{6}  W

e=\frac{151.228x10^{6} }{1000x10^{6} }*100= 15.12%

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4 years ago
Which type of communications equipment functions as a radio receiver and searches across several frequencies?
krok68 [10]

A <u>scanner</u> is a type of communications equipment that functions as a radio receiver and searches across several frequencies.

A scanner is a kind of a radio receiver that has the ability to receive multiple signals.

There are three modes which a scanner uses for acting as a radio receiver. The scan mode of the radio receiver constantly changes frequencies that helps in transmissions. There is also a manual scan mode that allows the users to search for their interested frequencies. The search mode allows the users to search through two sets of frequencies.

A scanner is a type of communication equipment that is easy to use with various features such as the volume, numeric keypad,  trunk tracking etc.

To learn more about scanners, click here:

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7 0
2 years ago
50 points Answer fast please
erma4kov [3.2K]

Answer:

22 hours

Explanation:

6 0
3 years ago
A box is hanging from two strings. String #1 pulls up and left, making an angle of 50° with the horizontal on the left, and stri
creativ13 [48]

Answer:

mb = 3.75 kg

Explanation:

System of forces in balance

ΣFx =0  

ΣFy = 0

Forces acting on the box

T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left

T₂  = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.

Wb :Weightt of the box (vertical downward)

x-y T₁ and T₂ components

T₁x= T₁cos50°

T₁y= T₁sin50°

T₂x= 30*cos75° = 7.76 N

T₂y= 30*sin75° = 28.98 N

Calculation of the Wb

ΣFx = 0  

T₂x-T₁x = 0

T₂x=T₁x

7.76 = T₁cos50°

T₁ = 7.76 /cos50° = 12.07 N

ΣFy = 0  

T₂y+T₁y-Wb = 0

28.98 + 12.07(cos50°) = Wb

Wb = 36.74 N

Calculation of the mb ( mass of the box)

Wb = mb* g

g: acceleration due to gravity = 9.8 m/s²

mb = Wb/g

mb = 36.74 /9.8

mb = 3.75 kg

8 0
3 years ago
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