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1 year ago

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What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti

ng on the slave cylinder? the master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter?

1 answer:

Nataliya [291]1 year ago

6 0

Archimedes principle states that

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

F2 = (πr2^2 / πr1^2) * mg

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

Substituting the values to the derived equation, we get

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

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A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m

ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

$S_{el} = 414Mpa$

Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

$E = \frac{414Mpa}{0.002}$

$E = 207Gpa$

Strain energy per unit volume at the elastic limit is

$U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}$

$U'_{el} = \frac{1}{2} (414)(0.002)$

$U'_{el} = 414kN\cdot m/m^3$

Considering that the net strain energy of the sample is

$U_{el} = U_{el}' \cdot (\text{Volume of sample})$

$U_{el} = U_{el}'(\frac{\pi d_0^2}{4})(L_0)$

$U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})$

$U_{el} = 2.663N\cdot m$

Therefore the net strain energy of the sample is $2.663N\codt m$

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1 year ago

Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

erica [24]

Answer:

Two possible points

<em>x= 0.67 cm to the right of q1</em>

<em>x= 2 cm to the left of q1</em>

Explanation:

<u>Electrostatic Forces</u>

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$

Assuming the positive sign :

$\displaystyle 0.02-x= 2x$

$\displaystyle 3x=0.02$

$\displaystyle x=0.00667\ m$

$x=0.67\ cm$

Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative

$\displaystyle 0.02-x=-2x$

$\displaystyle x=-0.02\ m$

$\displaystyle x=-2\ cm$

The negative sign of x means q3 is located to the left of q1 (assumed in the origin).

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2 years ago

An arthroscope can be inserted into a knee joint through a very tiny incision to help a doctor examine the knee. A light is tran

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Sample Response: The technology being described is fiber optics. Since they are small and flexible, they enable doctors to see areas that they might otherwise be unable to see without surgery.

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1 year ago

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Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic

Alinara [238K]

The volume of the room is the product of its dimensions:

$10\times 12 \times 4 = 480\text{ m}^3$

Now, from the equation

$d=\dfrac{m}{V}$

where d is the density, m is the mass and V is the volume, we deduce

$m=dV$

So, multiply the density and the volume to get the mass of air in the room.

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2 years ago

Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th

Inessa [10]

Answer:

F = 2.01*10^-16N -^k

Explanation:

In order to calculate the magnetic force perceived by the bee, you use the following formula:

$F=qv\ X\ B$ (1)

q: charge of the bee = 1pC = 1*10^-12 C

The average speed of a bee and the magnetic field of the earth are:

v = 6.70m/s

B = 30*10^-6 T

The bee is flying to the west (-^i). You consider that the magnetic field direction is to the north (^j). Then, the direction of the magnetic force is:

-^i X ^j = -^k

You replace the values of the parameters in the equation (1), in order to calculate the magnitude of the force:

$|F|=qvB=(1*10^{-12}C)(6.70m/s)(30*10^{-6}T)=2.01*10^{-16}N$

The magnetic force perceived by the bee is 2.01*10^-16N in the -^k direction, that is, toward the ground

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2 years ago