5. Electromagnetic radiation travels at a speed of 3.0 x 10^8 m/s.

A vector must always have both size and ..

DochEvi [55]

Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up

5 0

2 years ago

Who showed that our universe is heliocentric—the planets of the solar system revolve around the sun? Johannes Kepler Isaac Newto

denis23 [38]

Answer:

Option (3)

Explanation:

Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.

He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.

This theory describes the present-day motion of the planets around the sun in the solar system. This theory replaced the geocentric theory.

Thus, the correct answer is option (3).

6 0

2 years ago

1. The statement about Newton's reflecting telescope contains 1 error. Find the error, write it, and

Jlenok [28]

Answer:

i. The error is the rough convex mirror.

ii. This should be replaced with a smooth convex morror.

Explanation:

Reflection is dependent on the surface involved and has two types; diffuse and specular. When the surface is rough, diffused reflection is observed. The surface causes a distortion of the incident light (the rays would be reflected at different angles to one another) after reflection. This makes some rays to interfere with one another. While specular reflection is observed with a smooth surface.

In the statement, the rough convex mirror would produce a distorted reflection which would produce diffused reflection. The effect is that few or no rays (depending on the degree of how rough the surfce is) would be reflected to the other smooth, flat diagonal mirror.

8 0

2 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

The Coulomb force between two charges q1 and q2 at separation r in air is F. If half of the separation is filled with medium of

vodomira [7]

Answer:

The new Coulomb force is q₁q₂/9πε₀r²

Explanation

The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².

Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.

Since we have air between q₂ and q₃, the coulomb force between them is

F' = q₂q₃/4πε₀d²

= q₂(q₁/κ)/4πε₀(r/2)²

= 4q₂q₁/κ4πε₀r²

= 4/κ(q₂q₁/4πε₀r²)

= 4/9 × (q₂q₁/4πε₀r²)

= q₁q₂/9πε₀r²

So, the new Coulomb force is q₁q₂/9πε₀r²

3 0

3 years ago