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Deffense [45]
3 years ago
9

what is the point called at which the reflected rays intersect in the image above? a. real image b. virtual image c. focal point

d. optical axis
Physics
2 answers:
34kurt3 years ago
8 0
Focal point is the point where all of the reflected rays intersects. 
padilas [110]3 years ago
3 0

Answer:

A) Real Image

Explanation:

The point at which all reflected waves intersects is known as real image while if all the reflected ways seems to intersect at a point after reflection then that point is known as virtual image.

So converging type of light rays after reflection always forms real image while the light rays of diverging type is known as virtual image

so here correct answer will be

a) Real Image

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Explanation:

<h3>Hope it helps you!!</h3>

5 0
2 years ago
A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c
Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}

hence, the speed of the spheres is 37.45 m/s

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3 years ago
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alekssr [168]
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satela [25.4K]

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