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natita [175]
3 years ago
13

Find the resultant gravitational force exerted on the object at the origin by the other two objects. The universal gravitational

constant is 6.672 × 10−11 N · m2 /kg2 . Answer in units of N.
Physics
1 answer:
VladimirAG [237]3 years ago
7 0

Answer:

The resultant gravitational force is 7.76\times10^{-11}\ N

Explanation:

Suppose A coordinate system is constructed on the surface of a pool table, and three objects are placed on the coordinate system as follows: a 1.2 kg object at the origin, a 3 kg object at (0 m,1.8 m), and a 4.6 kg object at (4 m,0 m).

We need to calculate the gravitational force along x axis

Using formula of gravitational

F_{1}=\dfrac{GmM}{r^2}

Where, m = mass of first object

M = mass of object when placed at center

r = distance

G = gravitational constant

Put the value into the formula

F_{1}=\dfrac{6.672\times10^{-11}\times1.2\times4.6}{(4)^2}

F_{1}=2.301\times10^{-11}\ N

We need to calculate the gravitational force along y axis

Using formula of gravitational

F_{2}=\dfrac{Gm_{1}M}{r^2}

Put the value into the formula

F_{2}=\dfrac{6.672\times10^{-11}\times1.2\times3}{(1.8)^2}

F_{2}=7.413\times10^{-11}\ N

We need to calculate the resultant gravitational force

Using formula of resultant gravitational force

F_{net}=\sqrt{(F_{1}^2+F_{2}^2)}

F_{net}=\sqrt{(2.301\times10^{-11})^2+(7.413\times10^{-11})^2}

F_{net}=7.76\times10^{-11}\ N

Hence, The resultant gravitational force is 7.76\times10^{-11}\ N

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When a current flows in an aluminum wire of diameter 2.91 mm 2.91 mm , the drift speed of the conduction electrons is 0.000191 m
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Answer:

Number of electrons are flowing per second is 2.42 x 10¹⁹

Explanation:

The electric current flows through a wire is given by the relation :

I=envA   ....(1)

Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.

But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,

I=Ne      ....(2)

Here N is the rate of electrons passing through junction.

From equation (1) and (2).

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But area of wire, A=\pi \frac{d^{2} }{4}

Here d is diameter of wire.

So, N = nv\pi \frac{d^{2} }{4}

Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.

N = 6\times10^{28}\times 0.000191\times\pi \frac{(2.91\times10^{-3} )^{2} }{4}

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