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2 years ago

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A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field o

f magnitude . What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards

1 answer:

Ganezh [65]2 years ago

5 0

The question is incomplete. The complete question is :

A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.

Solution :

Given :

Length of the wire, L = 0.6 m

Mass of the wire length, m = 11 g

= $11 \times 10^{-3}$ kg

Magnetic field , B = 0.4 T

Know we know that :

ILB = mg

or $I=\frac{mg}{BL}$

$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$

$I=0.44963\ A$

$I = 449.63 \ mA$

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2 years ago

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Dao makes a table to identify the variables used in the equations for centripetal acceleration. A 2 column 5 rows. The first col

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Answer:

Column X. Tangential Speed

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Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

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Answer:

The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

$v=\dfrac{dx}{dt}$ ....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

$a=\dfrac{dv}{dt}$ ....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

$F=\dfrac{d(p)}{dt}$

Put the value of p

$F=\dfrac{d(mv)}{dt}$

$F=m\dfrac{dv}{dt}$

Put the value from equation (II)

$F=ma$

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

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2 years ago

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

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Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

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Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

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Calculate the mass of the air contained in a room that measures 2.50 m x 5.50 m x 3.00 m if the density of air is 1.29 g/dm3.53.

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Answer:

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Now we can find the mass of the air by using

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Substituting,

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2 years ago