All categories

3 years ago

How much pressure is applied to the ground

Physics

1 answer:

statuscvo [17]3 years ago

7 0

Answer:

Pressure applied by the man= 285103.125 $Pa$ or 41.35 $lb/in^{2}$

Explanation:

Pressure is defined as the perpendicular force applied per unit area.

i.e. $Pressure=\frac{Force}{Area}$

Now, $Force= mg$

where, $m$ = mass of the body(man) = 93 kg

$g$ = acceleration due to gravity of Earth = 9.81 $m/{s^{2}}$

$Area$ covered is equal to the area of both stilts(a man generally stands on two feet)

therefore $Area=2(0.04)^{2}$ $m^{2}$

and putting in the values, we get,

$Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}$

Now we need to convert to our required units:

$1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}$

(We can get the above result by individually converting kg to lb and meters to inches respectively)

Using the above relations we get,

$Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}$

You might be interested in

7. All of the following descriptions are true about descriptive method of research except one:

kolezko [41]

Answer:

Explanation:

3 0

3 years ago

A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,

Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. 

An initial condition has been supplied by the problem:

s=1.80m when t=0.245s

This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2)

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation):

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2

v=8.549m/s

Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t)

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like

0=(v)+(a)(t)

Substituting and solving for t:

0=(8.549m/s)+(-9.81m/s^2)(t)

t=0.8714s

Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2)

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window:

3.725m – 1.8m=1.925m

Answer:

1.925m

3 0

3 years ago

The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is

garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom = a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the water = 1000 kg/m³

The total pressure at the bottom of the pool = (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a = P A

ρ g h a = P A - P a

$h=\dfrac{P ( A-a)}{\rho g a}$

$h=\dfrac{P ( 1.5 a-a)}{\rho g a}$

$h=\dfrac{P ( 1.5- 1)}{\rho g}$

$h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m$

h=5.15 m

The depth is 5.15 m.

7 0

3 years ago

A substance that cannot be broken into simpler elements ia an

Anastaziya [24]

Element. element cannot be simplerfide

3 0

3 years ago

A classmate is attempting to draw a free body diagram for a box being pushed across the floor at a constant speed. The image bel

aivan3 [116]

Answer:

A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

A. The applied force should be the same size as the friction force

7 0

3 years ago