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IrinaVladis [17]
3 years ago
11

A person weighs 60 kg. The area under the foot of the person is 150 cm2. Find the pressure exerted on the ground by the person.

(Take g 10m/s2)
Physics
1 answer:
9966 [12]3 years ago
3 0

Answer:

40000 N/m²

Explanation:

Applying,

P =  F/A................... Equation 1

Where P = Pressure, F = Force, A = Area.

From the question,

The force(F) exerted by the person's foot is thesame as it's weight.

F = W = mg............ Equation 2

Where m = mass of the person, g = acceleration due to gravity.

Substitute equation 2 into equation 1

P = mg/A................ Equation 3

Given: m = 60 kg, g = 10 m/s², A = 150 cm² = (150/10000) m² = 0.015 m²

Substitute these values into equation 3

P = (60×10)/0.015

P = 600/0.015

P = 40000 N/m²

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Consider two uniform solid spheres where both have the same diameter, but one has twice the mass of the other. how much larger i
egoroff_w [7]
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5 0
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To determine the height of a tall building such as Sears Tower in Chicago, Illinois a ball was dropped from the top of the build
Darya [45]

Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3>y=v_{o} *t+\frac{1}{2} *g*t^{2}</h3><h3>y: The vertical distance the ball moves at time t  </h3><h3>v_{o}i: Initial speed </h3><h3>g=Gravity acceleration=9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )</h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s  is equal to height of Sears Tower (h).  

Too we know that the ball is released from rest, then,v_{0}=0

Height of Sears Tower calculation:

We replace  in the equation 1 the data following;

y=h

v_{o} =0

g=32,1\frac{ft}{s^{2} }

t= 9,5s

h=0*9.5+\frac{1}{2} *32.1*9.5^{2}

h=1448.5 ft

Answer: The height of Sears Tower is 1448.5 ft

6 0
3 years ago
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

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And the mass flow rate is  

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Since, the time and area is constant, the inlet and outlet will be same as

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As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

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is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

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As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

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So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

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