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Romashka-Z-Leto [24]
4 years ago
15

Explain the difference in pressure between a small and a large point

Physics
1 answer:
Radda [10]4 years ago
5 0

Answer:

If the force applied is the same, the small point will exert a greater pressure. Assumption:

  • both forces are applied perpendicularly and evenly, and
  • the entirety of the larger point is in contact with the object that it is acting on.  

Explanation:

Pressure is the amount of force exerted per unit area. In other words, if a normal force of size N is exerted over an area of A, it will exert a pressure of

\displaystyle P = \frac{N}{A}.

The contact area, A, is in the denominator. If the numerator stays the same and the value of the denominator increases, the value of the quotient will become smaller. With a similar logic, if the value of the numerator (normal force) stays the same, raising the size of the force (the denominator) will reduce the pressure on the object.

This relationship is widely applied in real life. For example, the moon soil is known to be soft; it is likely to get deshaped easily under pressure. Landers with pointy legs might sink into the soil and topple. To minimize that risk, each landing leg of the Apollo Lunar Modules is equipped with a wide plate, The plate maximizes the contact area and minimize the pressure on the lunar surface.

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Jim is driving a 2268-kg pickup truck at 19 m/s and releases his foot from the accelerator pedal. The car eventually stops due t
lord [1]

Answer:

The initial kinetic energy of the truck is 409374 J

Explanation:

This problem can be solved in two ways. Let´s solve it first in the easiest way.

The kinetic energy is calculated using this equation:

E = 1/2 · m · v²

Where:

E = kinetic energy

m = mass

v = velocity

Then, the kinetic energy of the truck will be:

E = 1/2 · 2268 kg · (19 m/s)² = 409374 J

And that´s it.

But we can complicate it a bit:

The kinetic energy is the work needed to move an object from rest to a desired velocity. If the object is moving, the work needed to stop it must be of the same magnitude as its kinetic energy (in the opposite direction to the movement).

The equation for work is:

W = F · d

Where:

W= work

F = force

d = distance

We know the magnitude of the force applied to the truck, but we do not know for how much distance that force was applied. The distance can be calculated using the equation for the position of an object moving in a straight line:

x = x0 + v0 · t + 1/2 · a · t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

But we still do not know the time nor the acceleration.

The acceleration can be obtained from the equation of force:

F = m · a

Where

F = force

m = mass

a = acceleration

Then:

900 N = 2268 kg ·a

a = 900 N /2268 kg = 0.397 m/s²

Now, we can calculate the time needed for the truck to stop. We know that at the final time, the velocity is 0. Then, we can use the equation for velocity to obtain that time:

v = v0 + a · t

Where:

v = velocity at time t

v0 = initial velocity

a = acceleration

t = time

Then:

v = 19 m/s - 0.397 m/s² · t

0 = 19 m/s - 0.397 m/s² · t

-19 m/s / -0.397 m/s² = t (acceleration is negative because it is opposite to the direction of the movement)

t = 47.86 s (The truck stoped at 47.86 s after releasing the foot from the accelerator pedal)

With the time and acceleration, we can calculate the traveled distance.

x = 0 m + 19 m/s · 47.86 s - 1/2 · 0.397 m/s² · (47.86s)²

x = 454.66 m (without rounding the acceleration nor the time, the value will be 454.86 m)

Now, we can calculate the work done to stop the truck which will be of the same magnitude as the kinetic energy:

W = 900 N · 454.66 m = 409194 J

(if you do all the calculations without rounding, you will get the same value as we calculated above using the equation of kinetic energy, 409374 J).

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