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Gwar [14]
2 years ago
15

An ideal gas in a cylinder occupies a volume of 0.065 m3 at room temperature (T = 293 K). The gas is confined by a piston with a

weight of 100 N and an area of 0.65 m2. The pressure above the piston is equal to one atmosphere (atm = 1.013x105 Pa). The piston is free to move up and down. 1) What is the magnitude of the net force on the piston? zero 50 N 100 N 2) Briefly explain your answer.
Physics
1 answer:
slamgirl [31]2 years ago
8 0

Answer:

zero

Explanation:

There are three forces acting on the piston

1. force due to atmospheric pressure = F1 downward

2. force due to gaseous pressure = F2 upward

3. force due to the weight placed on the piston = F3 = mg downward

As the piston is in equilibrium condition, so the net force on the piston is zero.

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dmitriy555 [2]
30 seconds is the answer
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2 years ago
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) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force
Nana76 [90]
Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others. 
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
F_x=F_3+F_2cos(45^{\circ})
F_y=F_1+F_2sin(45^{\circ})
Let's figure out each component:
F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}
Total force acting on the charge would be:
F=\sqrt{F_x^2+F_y^2}
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N
F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N
Now we can find the total force acting on a single charge:
F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N
As said before, intensity of the force acting on charges is the same for all of them.

5 0
2 years ago
Sachi wants to throw a water balloon to knock over a target and win a prize. The target will only fall over if it is hit with a
posledela

Answer:

3.1 m/s²

Explanation:

Given:

Mass of the balloon (m) = 11.4 g = 0.0114 kg ( 1 kg = 1000 g)

Force acting on the balloon (F) = 0.035 N

Acceleration with which the balloon must be hit (a) = ?

Now, we know that, from Newton's second law, net force acting on an object is equal to the product of its mass and acceleration.

Therefore, framing in equation form, we have:

F=ma

Rewriting in terms of acceleration 'a', we get:

a=\frac{F}{m}

Now, substitute the given values and solve for 'a'. This gives,

a=\frac{0.035\ N}{0.0114\ kg}\\\\a=3.07\approx 3.1\ m/s^2(Nearest\ tenth)

Therefore, the acceleration of the water balloon to reach the target must be equal to 3.1 m/s².

7 0
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Randy wants to know whether a soil's porosity affects how easily seedlings grow in it.
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he plants seedlings in soils with different levels of porosity and equal levels of permeability. 

Permeability is not what needs to be tested. If it changes, you may not be able to determine whether it was the porosity or permeability that cause changes.  
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2 years ago
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