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AfilCa [17]
3 years ago
13

A gas sample is confined within a chamber that has a movable piston. A small load is placed on the piston; and the system is all

owed to reach equilibrium. If the total weight of the piston and load is 70.0 N and the piston has an area of 5.0 × 10 –4 m 2 , what is the pressure exerted on the piston by the gas? Note: Atmospheric pressure is 1.013 × 10 5 Pa.
Physics
1 answer:
timurjin [86]3 years ago
7 0

Explanation:

It is given that,

Total weight of the piston, W = F = 70 N

Area of the piston, a=5\times 10^{-4}\ m^2

Let P is the pressure exerted on the piston by the gas. The force per unit area is called the pressure exerted pressure of the gas. Mathematically, it is given by :

P=\dfrac{F}{A}

P=\dfrac{70\ N}{5\times 10^{-4}\ m^2}

P=1.4\times 10^5\ Pa

We know that the atmospheric pressure is given by :

P_o=1.013\times 10^5\ Pa

So, the pressure is given by :

p=P+P_o

p=1.4\times 10^5+1.013\times 10^5

p=2.41\times 10^5\ Pa

Hence, this is the required solution.

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Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and
kow [346]

A) Acceleration: a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

A)

Acceleration is defined as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first  segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

a_1 = \frac{11-8}{4}=0.75 m/s^2 (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: a_2 = 0

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

a=\frac{0-11}{7}=-1.57 m/s^2

And the negative sign means the acceleration is south, opposite to the direction of motion.

B)

In a uniformly accelerated motion, the displacement can be calculated as:

s=ut+\frac{1}{2}at^2

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

u = 0\\a = 0.75 m/s^2\\t=4 s

So the displacement is

s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m

- For the second segment, we have

u = 11 m/s\\a = 0\\t=15 s

So the displacement is

s_2 = (11)(15)+0=165 m

- For the third segment, we have

u = 11\\a = -1.57 m/s^2\\t=7 s

So the displacement is

s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

C)

The average velocity is given by:

v=\frac{d}{t}

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

v=\frac{209.5}{26}=8.06 m/s (north)

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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7 0
3 years ago
Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Offic
miss Akunina [59]

Answer:

0.256 hours

Explanation:

<u>Vectors in the plane </u>

We know Office A is walking at 5 mph directly south. Let X_A be its distance. In t hours he has walked

X_A=5t\ \text{miles}

Office B is walking at 6 mph directly west. In t hours his distance is

X_B=6t\ \text{miles}

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

D=\sqrt{X_A^2+X_B^2}

D=\sqrt{(5t)^2+(6t)^2}

D=\sqrt{61}t

This distance is known to be 2 miles, so

\sqrt{61}t=2

t =\frac{2}{\sqrt{61}}=0.256\ hours

t is approximately 15 minutes

3 0
3 years ago
URGENT!! <br>Question in attachment. <br>thanks ​
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Answer:9

Explanation:beacause

6 0
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What type of plate boundary decreases the amount of the Earth's crust?
rosijanka [135]

Answer:

Convergent.

Explanation:

Just as oceanic crust is formed at mid-ocean ridges, it is destroyed in subduction zones. Subduction is the important geologic process in which a tectonic plate made of dense lithospheric material melts or falls below a plate made of less-dense lithosphere at a convergent plate boundary.

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2 years ago
a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl
bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

s=ut+\dfrac{1}{2}gt^2

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

230=0+\dfrac{1}{2}\times9.8\times t^2

t=\sqrt{\dfrac{230\times2}{9.8}}

t=6.85 sec

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

7 0
3 years ago
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