ii) Telescopes uses two convex lenses producing a magnified image while the human eye only possesses one convex lens (image seen are smaller than that viewed under telescopes)

Explanation:

The telescopes can be used to view far distant objects due to their presence of two convex lenses. The two convex lenses are the objective lens (lens closer to object) and the eye piece lens (lens closer to eye). The object to be viewed forms an intermediate image first before the final image is seen using the eye piece lens.

The human eye only possess one convex lens and as such cannot view far ranged objects.

8 0

3 years ago

Suppose the displacement of an object is related to time according to the expression x=By*2, what are the dimensions of B

laiz [17]

Answer:

Explanation:

AP3X

5 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

A magnet is moved in and out of a coil of wire connected to a high-resistance voltmeter. If the number of coils doubles, the ind

Varvara68 [4.7K]

Answer:

A. Doubles.

Explanation:

In an electromagnetic device such as a generator, when a wire (conductor) moves through the magnetic field between the South and North poles of a magnet, an electromotive force (e.m.f) is usually induced across a wire

The mode of operation of a generator is that a metal core with copper tightly wound to it (conductor coil) rotates rapidly between the two (2) poles of a horseshoe magnet type. Thus when the conductor coil rotates rapidly, it cuts the magnetic field existing between the poles of the horseshoe magnet and then induces the flow of current.

When a high-resistance voltmeter is connected to an electric circuit, a deflection will arise due to the flow of electricity. Moving the magnet towards the coil of wire will cause the needle of the high-resistance voltmeter to move in one direction. Also, as the magnet is moved out from the coil of wire, the needle of the high-resistance voltmeter moves in the opposite direction.

In this scenario, a magnet is moved in and out of a coil of wire connected to a high-resistance voltmeter. If the number of coils doubles, the induced voltage doubles because the number of turns (voltage) in the primary winding is directly proportional to the number of turns (voltage) in the secondary winding.

4 0

3 years ago