a piston-cylinder apparatus has a 4 kg piston with a diameter of 5 cm resting on a body of water 60 cm high. atmospheric pressur
1 answer:
The mass of the water in the container is 10,104.00 kg.
A schematic diagram of a piston-cylinder apparatus is shown in the adjoining image.
In the above case, the atmospheric pressure applied by the piston and the weight of the piston should balance the weight of the water inside the container
Hence,
Atmospheric Pressure x Area of Piston + Mass of piston x g = Mass of water x g
101 x 10^3 x pi x (2.5 x 10^-2)^2 + 4 x 10 = Mass of water x 10
Mass of water x 10 = 101,040.001
Mass of Water = 101,040.001/10
Mass of Water = 10,104 kg
Hence, the mass of the water in the container is 10,104.00 kg.
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Answer:
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω
Explanation:
Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.
When the current increases by 0.450 A wen only R1 is in the circuit, the current is
9/R1 + R2 + 0.450 A = 9/R1 (1)
When the current increases by 0.225 A when only R2 is in the circuit, the current is
9/R1 + R2 + 0.225 A = 9/R2 (2)
equation (1) - (2) equals
9(1/R1 - 1/R2) = 0.450 A - 0.225
9(1/R1 - 1/R2) = 0.125
(1/R1 - 1/R2) = 0.125 A/9 = 0.0138
1/R1 = 0.0138 + 1/R2
R1 = R2/(1 + 0.0138R2) (3)
From (1)
9/R1 - 9/R1 + R2 = 0.450 A
9R2/[R1(R1 + R2)] = 0.450 A
R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5
R2/[R1(R1 + R2)] = 0.5 (4)
From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,
(1 + 0.0138R2) = 0.5(R1 + R2)
0.5R1 + 0.5R2 = 1 + 0.0138R2
0.5R1 = 1 + 0.0138R2 - 0.5R2
0.5R1 = 1 - 0.4862R2 (5)
Substituting (3) into (5) we have
0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2
R2 = (1 + 0.0138R2)(1 - 0.4862R2)
R2 = 1 - 0.4724R2 - 0.0067R2²
Collecting like terms, we have
0.0067R2² + 0.4724R2 + R2 - 1 = 0
0.0067R2² + 1.4724R2 - 1 = 0
Using the quadratic formula,
We choose the positive answer.
So R2 = 0.340 Ω
From (5)
R1 = 0.5 - 0.9931R2
= 0.5 - 0.9931 × 0.340
= 0.5 - 0.338
= 0.162 Ω
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω
Answer:
20.4
Explanation:
To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.
First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.
Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.
Applied force = 1055 N (490 + 565)
Friction force= Unknown
To find the friction force, use the kinetic friction formula, Friction = μkN
μk is the coefficient, which the problem includes- it is 0.613.
N is the normal force, which we have to find.
*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.
Weight = mg = (40)(9.8) = 392 N
Normal force = 392 N
Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N
Friction = 240.296 N
Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N
Horizontal Net Force: 814.704 N
Now that we know the net force, plug in the numbers for the formula
∑F= ma.
814.704 = (40.0)(a)
*Divide on both sides)
a = 20.3676 m/s^2
Round it to 3 significant figures, to get:
20.4
