Unlike mechanical waves such as sound or earthquake waves, EM waves can travel through empty space. True or false

Hardness refers to the strength of the forces holding atoms together in a solid mineral. The Mohs scale will tell how easily a m

Vikki [24]

Answer:

Determine how rare a diamond is

5 0

3 years ago

Find the work done by a man who is pulling a box of 45kg of mass by means of rope which makes angle of 45 degrees.

Juli2301 [7.4K]

Answer:

0

Explanation:

Since no distance is given, the force is not doing any work

No work is done by the man since we do not know the distance or displacement.

Work is only said to be done when the force applied on an object moves it through a particular distance.

Work done = Force x distance.

Since no distance is given in this problem, we can as well assume that the force applied is doing no work on the object.

4 0

3 years ago

The amplitude of the electric field for a certain type of electromagnetic wave is 570 N/C. What is the amplitude of the magnetic

vesna_86 [32]

Given Information:

Amplitude of the electric field = E = 570 N/C

Speed of light = c = 3.0x10⁸ m/s

Required Information:

Amplitude of the magnetic field = ?

Answer:

Amplitude of the magnetic field = 1.90x10⁻⁶ T

Explanation:

The relation between the amplitude of the electric field and magnetic field is given by

B = E/c

Where c is the speed of light

B = 570/3.0x10⁸

B = 1.90x10⁻⁶ T

Therefore, the amplitude of magnetic field is 1.90x10⁻⁶ T

8 0

4 years ago

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

3 years ago

The pitch of sound is related to the ___ of the sound waves

solniwko [45]

Answer:

The answer is frequency - I think....sorry if I'm wrong.

4 0

3 years ago

Read 2 more answers