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miskamm [114]
3 years ago
8

A mass of 10 g of oxygen fill a weighted piston–cylinder device at 20 kPa and 110°C. The device is now cooled until the tempe

rature is 0°C. Determine the change of the volume of the device during this cooling.
Physics
1 answer:
mezya [45]3 years ago
6 0

Answer:

The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

\Delta V=V_{2}-V_{1}

\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})

Put the value into the formula

\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)

\Delta V=14.297\ L

\Delta V=14.3\times10^{-3}\ m^3

Hence, The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

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A rectangular block of steel measures 4cm*2cm*1.5cm*. its mass is 93.6g
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Answer:

b

Explanation:

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3 years ago
You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

F_f = \mu m g

So the acceleration of car due to friction is given as

F_{net} = - \mu mg

a = \frac{F_{net}}{m}

a = -\mu g

now it is given that

\mu = 0.868

g = 9.81 m/s^2

so here we have

a = -0.868 * 9.81

a = -8.52 m/s^2

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0
3 years ago
What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of?
madreJ [45]
We will apply the Newton's second Law so the we will be able to find the acceleration.
F (tot) = ma
a = F(tot) /  m
a = 32.0 N / 65.0 kg = 0.492 m/s^2
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7 0
3 years ago
Does anyone know this
NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net  force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the  right

Explanation:

1)

To find the net force, we have to analyze all the forces acting on the box.

We have:

  • Force to the right: F_a = 20 N, the applied force
  • Force to the left: F_f = 4 N, the force of friction
  • Force to the bottom: F_g = 400 N, the weight of the box (the weight is always downward vertically)
  • Force to the top: F_N = 400 N. This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

F_x = F_a - F_f = 20 - 4 = 16 N (to the right)

While the vertical force is

F_y = F_N - F_g = 400 - 400 = 0

So the net force is 16 N to the right.

2)

In this case, we have the following forces:

  • F_g = 4 N downward, the weight of the ball
  • F_a = 100 N to the left, the force that kicks the ball
  • F_f = 2 N to the right, the force of friction
  • F_N = 4 N upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

F_x =F_a - F_f = 100 -2 = 98 N (to the left)

While the vertical force is

F_y = F_g - F_N = 4 - 4 = 0 (downward)

And so, the net force is 98 N to the left.

3)

The force acting on the squirrel in this problem are:

  • F_g = 8 N downward, the weight of the squirrel
  • F_f = 7.5 N upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

F_y = F_g - F_f = 8 - 7.5 = 0.5 N

And since the weight is larger than the air resistance, the direction of the net force is downward.

4)

The forces acting on Monkey are:

  • F_1=95 N is the force applied to the right by Bunny
  • F_2 = 75 N is the force applied by Deer from the left (so, also on the right)
  • F_g = 50 N is the weight of Monkey, downward
  • F_N = 50 N is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

F_x = F_1 + F_2 = 95+75=170 N (to the right)

While the net force in the vertical direction is

F_y = F_N - F_g = 50 - 50 = 0

And therefore the net force is 170 N to the right

5)

The forces acting on Deer are:

  • F_a = 100 N + 100 N = 200 N to the right, the combined force applied by Bunny and Monkey
  • F_f = 25 N to the left, the force of friction
  • F_g = 150 N downward, the weight of the deer
  • F_N = 150 N upward, the normal reaction from the surface that balances the weight

So the net horizontal force is

F_x = F_a - F_f = 200 - 25 = 175 N to the right

While the net vertical force is

F_y = F_N - F_g = 150 - 150 = 0

So the net force is 175 N to the right.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

3 0
3 years ago
A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with
romanna [79]

Answer:

The correct answer to the following question will be "41.87 m".

Explanation:

The given values are:

The speed of trooper = 30.1 \ m/s

The velocity of red car = 45.4 \ m/s

Now,

A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at

t=\frac{45-30}{2.7}                                                              (∵ time=\frac{distance}{time})

t=5.56 \ sec

then,

The distance covered by trooper,

t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2

   =208.33 \ m

The distance covered by red car,

= 45\times 5.56

= 250.2 \ m

Maximum distance = 250.2-208.33

                                = 41.87 \ m                                                        

6 0
3 years ago
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