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3 years ago

For a converging lens, a ray arriving parallel to the optic axis

Physics

1 answer:

mixer [17]3 years ago

8 0

Answer:

b. passes through the principal focal point.

Explanation:

Light wave can be defined as an electromagnetic wave that do not require a medium of propagation for it to travel through a vacuum of space where no particles exist.

A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.

Basically, there are two (2) main types of lens and these includes;

I. Diverging (concave) lens.

II. Converging (convex) lens.

A converging lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. This type of lens is usually thin at the lower and upper edges and thick across the middle.

For a converging lens, a ray arriving parallel to the optic axis passes through the principal focal point.

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Consider the concepts of kinetic energy (KE) and gravitational potential energy (GPE) as you complete these questions. A ball is

Lena [83]

Answer:

When the ball is held motionless above the floor, the ball possesses only GPE energy.If the ball is dropped, its GPE energy decreases as it falls.If the ball is dropped, its KE energy increases as it falls.

Explanation:

If the ball is held motionless, then its kinetic energy is equal to zero, since kinetic energy depends on the velocity. And the ball is held above the ground, which means it possesses gravitational potential energy.

If the ball is dropped, its height will decrease, therefore its gravitational potential energy will decrease. Along the way, the ball will be in free fall, and therefore its velocity will increase, hence its kinetic energy.

$K = \frac{1}{2}mv^2\\U = mgh$

3 0

2 years ago

A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic

Mazyrski [523]

Answer:

The total resistance of the wire is = $1.917\times10^{-3}$

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, $R_{Total}$

$\frac{1}{R_{Total}} =\frac{1}{R_{cu} }+\frac{1}{R_{al}}$

Parameters given:

Length of wire = 1 m

Cross sectional area of copper $A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3} )^{2} =3.142\times10^{-6} m^{2}$

Cross sectional area of aluminium wire

$A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3} )^{2}-(1\times 10^{-3} )^{2}] =9.42\times10^{-6} m^{2}\\$

Resistivity of copper $\rho _{cu}=1.7\times 10^{-8} \Omega .m$

Resistivity of Aluminium $\rho _{al}=2.8\times 10^{-8} \Omega .m$

Resistance of copper $R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} } =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega$

Resistance of aluminium $R_{al}= \frac{\rho_{al} \times l}{A_{al} } =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega$

The total resistance of the wire can be obtained as follows;

$\frac{1}{R_{Total}} =\frac{1}{5.41\times10^{-3} }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}$

$R_{Total}= 1.917\times 10^{-3}\Omega$

∴ The total resistance of the wire = $1.917\times 10^{-3}\Omega$

4 0

2 years ago

Read 2 more answers

What's true about the elliptical path that the planets follow around the sun? A. A line can be drawn from the planet to the sun

wariber [46]

Answer:

A. A line can be drawn from the planet to the sun that sweeps out equal areas in equal times

Explanation:

This is exactly what Kepler's second law of planetary motion states:

"the segment joining the sun with the center of each planet sweeps out equal areas in equal time"

This law basically tells how the speed of a planet orbiting the sun changes during its revolution. In fact, we have that:

- when a planet is closer to the Sun, it will orbit faster

- when a planet is farther from the Sun, it will orbit slower

5 0

3 years ago

What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long water tank that can hold 50.0 kg of water by the weight

svetoff [14.1K]

Answer:

1088.9N/m2

Explanation:

Calculation for What pressure is exerted

First step is to find the area of bottom of the tank using formula

Area=Width*breadth

Let plug in the formula

Area=0.5*0.9

Area=0.45m2

Now let calculate what pressure is exerted using this formula

Pressure=Force/Area

Where,

Force=Mass *Gasoline

Area=Width of the tank* Length of the tank

Let plug in the formula

Pressure=50*9.8/0.5*0.9

Pressure=490/0.45

Pressure=1088.9N/m2

Therefore What pressure is exerted is 1088.9N/m2

3 0

2 years ago

two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t

Tomtit [17]

The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

$F=\frac{kq^2}{r^2}$

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

$q_1 = 6 \mu C\\q_2 = -6 \mu C$

Learn more about electric force:

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3 0

2 years ago