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3 years ago

Two uniform spheres,each of mass 0.260kg are fixed at points A and B

Physics

1 answer:

Ivan3 years ago

3 0

Answer:

a) 2.76x10^-9 m/s^2

b) downward

Explanation:

a) In order to do this, we need to use the newton's law of gravitation which is:

F = G.m/r²

Where:

G: Constant of gravitation (6.67x10^-11 N/m)

m: mass of the object

r: distance or radius of the object.

In this case, we have two spheres, so this law becomes:

F = G*m1*m2 / r²

Now, you are not providing a picture with this, however, this problem is very similar to one I solved before, so, I'm gonna use the same values to solve this. If, you have different values for the distance, just replace them in this procedure and should get the correct and more accurate answer.

In this case, the distance that separates sphere in P from sphere A and B is 10 cm. As both spheres have the same mass and they are at the same distance from P, we can assume that F1 = F2 so:

F1 = (6.67x10^-11) * (0.01) * (0.26) / (0.1)² = 1.73x10^-11 N

sphere at P, will roll downward to the middle of sphere A and B. The distance between these two points it's 8 cm (See picture attached)

Now, according to my data, they are exerting a force in the y axis, so:

F1y = Fy * senФ

senФ using trigonometry is:

senФ = opposite lenght / hipotenuse

senФ = 8/10 = 0.8

Replacing:

F1y = 1.73x10^-11 * 0.8 = 1.38x10^-11 N

This force it's exactly the same as F2y, so the total force is:

F = 1.38x10^-11 * 2 = 2.76x10^-11 N

Finally, using the second law of Newton:

F = m*a

Solving for a:

a = F/m

replacing:

a = 2.76x10^-11 / 0.01

a = 2.76x10^-9 m/s²

According to the picture and the value of this magnitude, it's going downward.

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

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3 years ago

What are the disadvantages of driverless cars? Check all that apply

Sauron [17]

Answer:

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2 years ago

When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink

Dovator [93]

Answer:

y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

Therefore the diffraction equation for slits with m = 1 remains

a sin θ = λ

in general these experiments occur for oblique angles so

sin θ = θ

θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

θ = s / R

as the angle is small the arc approaches the distance s = y

y / R = 1.22 λ / s

y = 1.22 λ R / a

let's calculate

y = 1.22 500 10⁻⁹ 0.42 / 0.032

y = 8 10⁻⁶ m

with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)

y <8 10⁻⁶ m

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3 years ago

Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?

Helen [10]

Answer:

A. Mass and acceleration

Explanation:

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Answer:

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