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3 years ago

This problem is based on the whole idea of pressure but I’m having trouble on when the area circle formula is included.

Physics

1 answer:

Mice21 [21]3 years ago

8 0

Answer:

6.23x10^6Pa

Explanation:

Data obtained from the question include:

F (force) = 490N

r (radius) = 0.005m

A (area of the circlular heel) =?

P (pressure) =.?

First, we'll begin by calculating the area of the circlular heel. This is illustrated below:

Area of circle = πr^2

Area = 22/7 x (0.00)^2

Area = 7.86x10^-5m^2

Pressure is simply force per unit area. It represented mathematically as

Pressure = Force /Area

Pressure = 490/7.86x10^-5

Pressure = 6.23x10^6N/m2

Recall: 1N/m2 = 1Pa

Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa

Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor

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vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

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m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
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Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

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