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3 years ago

An object is held at a distance of 20cm from a concave lens of focal length 80cm. What is the

Physics

1 answer:

Sladkaya [172]3 years ago

5 0

<h3>Answer</h3>

it is -16

<h2>explanation </h2><h3>Here, u=−20cm,f=−80cm,v=?,h1=2cm,h2=?</h3><h3>From 1v−1u=1f,1v=1f+1u=1−80−120=−1−480=−580</h3><h3>v=−805=−16cm.</h3><h3>Thus, image is at 16 cm from the lens on the same side as the object.</h3><h3>As h2h1=vu∴h2=vuh1=−16−20×2cm=1.6cm</h3><h3>This is the size of the image.</h3>

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Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

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