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3 years ago

The mean orbital radius of the earth around the sun 1.5 × 108 km. Calculate the

Physics

1 answer:

Readme [11.4K]3 years ago

5 0

Answer:

M = 1.994 × 10^(30) kg

Explanation:

We are given;

Orbital radius; r = 1.5 × 10^(8) km = 1.5 × 10^(11) m

Gravitational constant; G = 6.67 × 10^(-11) N.m²/kg²

If the orbit is circular, the it means the gravitational force is equal to the centripetal force.

Thus; F_g = F_c

GMm/r² = mv²/r

Simplifying gives;

GM/r = v²

M = v²r/G

Now, v is the speed of the earth around the sun and from online sources it has a value of around 29.78 km/s = 29780 m/s

Thus;

M = (29780^(2) × 1.5 × 10^(11))/6.67 × 10^(-11)

M = 1.994 × 10^(30) kg

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The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0

3 years ago

A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an

Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

R is resistance

The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

ΔV=IR

$=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V$

ΔV=0.484mV

3 0

3 years ago

A student pushes on a crate with a force of 100 Newtons directed to the right.

romanna [79]

Answer:

100N

Explanation:

Newton's third law of motion

For every action, there is an equal and opposite reaction.

Therefore 100N of force is exerted by the crate on student as a reaction to his action

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3 years ago

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Answer:

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Explanation:

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