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3 years ago

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Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires

are connected end to end, so that the resultant wire has length 10L. What is the Young's modulus of the resulting wireA) 0.1YB) YC) 10YD) 100Y

1 answer:

svlad2 [7]3 years ago

8 0

Answer:

option (B)

Explanation:

Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.

Its unit is N/m².

The formula for the Young's modulus is given by

$Y=\frac{F \times L}{A\times \delta L}$

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.

It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.

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The radius of the satellite is given [R] $2.5*10^{12} m$ .

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

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where T is the time period of the satellite and R is the radius of the satellite.

$V=\frac{2*3.14*10^{12} }{2*10^{6} }$

$= 7.85*10^{6} m/s$

There is also another way through which we can get the solution as explained below-

We know that the tangential speed of a satellite V $=\sqrt{\frac{GM}{R^{2} } }$

where G is the gravitational constant and M is the mas of central object.

But we know that $g=\frac{GM}{R^{2} }$

⇒ $GM=gR^{2}$ where g is the acceleration due to gravity of that central object.

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By knowing the value of g due to that central object we can also calculate its tangential speed.

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As we know that when charge particle is projected in perpendicular magnetic field then the radius of the charge particle is given as

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now we have

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since here radius of proton and electron will be same

so we will have

$r_e = r_p$

$\frac{m_e v}{q_e B_e} = \frac{m_p v}{q_p B_p}$

so we have

$B_e = \frac{B_p m_e}{m_p}$

given that

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$m_e = 9.11 \times 10^{-31} kg$

$m_p = 1.67 \times 10^{-27} kg$

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$B_e = \frac{0.50(9.11\times 10^{-31})}{1.67\times 10^{-27}}$

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a 905 - g meteor impacts the earth at a speed of 1623 m/s. if all of its energy is entirely converted to heat in the meteor, wha

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From law ofconservation of energy, we have that

1/2mv^2 = mcΔθ

By cancelling "m" on both sides, we have that

v^2/2 = cΔθ

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Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5

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Region 2; 2<r<4;

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