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3 years ago

Can a small force exert a greater torque than a larger force.Explain.

Physics

1 answer:

butalik [34]3 years ago

6 0

Answer:

Yes, a small force can exert a greater torque than a large if small force has large enough lever arm.

Explanation:

We know that torque exert on an object is given by

$\tau=Force\times perpendicular\;distance$

Torque is directly proportional to force

$\tau\propto F$

Torque is directly proportional to perpendicular distance

If force is small then torque exert is small

If force is large then torque exert is large

If lever arm is small then torque is small

If lever arm is large then torque is large.

A small force exert a greater torque than a large force if small force has large enough lever arm.

Yes, a small force can exert a greater torque than a large.

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A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, $\lambda=\dfrac{\lambda_ox_o}{x}$

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$ ..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

$E=\dfrac{k\lambda_o}{2x_o}$

Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

In your own words, explain what conservation of energy means. Also, give an example of the conservation of energy using somethin

Genrish500 [490]

Energy can not be created or destroyed but can change from one form to another.

example: as a roller coaster cart loses height the more speed it gains, the potential energy is transferred into kenetic energy

8 0

3 years ago

Your brother is insisting that you’ll never hear a sound produced behind a barrier wall at the end of your yard you notice that

Tresset [83]

Answer

D.Diffraction

Explanation

Diffraction is a property that is experienced by waves when they come across a barrier when they are in motion.

The ways tends to curve behind the barrier. This is called diffraction of waves.

Now, sound is a wave and it also experience diffraction. . So the brother will be able to hear the sound due to diffraction

8 0

3 years ago

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What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

6 0

3 years ago