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scZoUnD [109]
3 years ago
11

Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f

it inside. Block B has a bigger volume than block A. Which of the following is true?
A)Density of B is the same as Density of A


B)Density of B is greater than Density of A


C)Density of B is smaller than Density of A


D)Not enough information

Physics
1 answer:
RSB [31]3 years ago
3 0

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

The force exerted on each piston is just equal to the weight of the corresponding mass: F=W=mg, where m is the mass and g is the gravitational acceleration. So the equation becomes

\frac{m_A g}{A_A}=\frac{m_B g}{A_B}

Now we can rewrite the mass as the product of volume, V, times density, d:

\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}

We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

So we can further re-arrange the equation (and simplify g as well):

\frac{V_A d_A}{1}=\frac{V_B d_B}{2}

\frac{d_A}{d_B}=\frac{V_B}{2V_A}

We are also told that block B has bigger volume than block A: V_B > V_A. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

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Assuming you mean how they work,

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By what factor would your weight change if the earth's diameter were doubled and its mass were also doubled?
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Aha! This is a fun question, likely AP Physics 1 if I'm not mistaken. This is how you solve it, you need the following equation.

F=G\frac{m_{1} *m_{2}}{r^2}

Now on this question you don't need to actually use the mass and diameter of the earth as we are only looking at the factor of how much your weight would change, so you can just plug in some random numbers for the mass and radius. In my case I will use the original radius as being 2 meters and the original mass of the earth as being 20kg. You can actually just omit the mass of your body and the gravitational constant isn't necessary to be multiplied by because they will remain the same in both scenarios.

Original weight:

F= \frac{20kg}{2^2m}
F=5N

So the original weight in this circumstance is 5N. Now then if we double the mass to 40kg and the radius to 4m you will have the following.

F= \frac{40kg}{(4)^2m}
F=2.5N

So the original weight was 5N and after doubling the mass and radius the weight reduced to 2.5N. This means the factor which your weight would change by is .5 or decrease by 1/2.

I hope this helps you :) You can ask me if you have any additional physics questions.
7 0
2 years ago
A wire is wrapped multiple times around a galvanized or aluminum nail and then each end of the wire connected to a battery for s
kiruha [24]
<span>A)the paper clip is repelled away from the nail because an electromagnetic field magnetized the nail</span>
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If the first maximum of one circular diffraction pattern passes through the center of a second diffraction pattern, the two sour
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Answer: option A is correct

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A satellite of mass m circles the earth a distance R from the center of the earth.If the radius of the earth is 6.0*10^6ms.calcu
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(a) The height above the earth's surface of the parking orbit is determined from the difference between the orbital radius and earth's radius (h = r - R).

(b) Velocity of the satellite of the earth is determined with a formula given as  (2π/T)r.

<h3>Height above the earth's surface </h3>

The height above the earth's surface is calculated using the following formula;

T = \frac{2\pi r^{\frac{3}{2} }}{\sqrt{Gm} }

where;

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r = h + R

h = r - R

where;

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The velocity of the satellite is determined using the formula below;

v = ωr

v = (2π/T)r

Learn more about velocity of satellite here: brainly.com/question/22247460

#SPJ1

5 0
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