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3 years ago

Which of the following is an electromagnetic wave?

Physics

2 answers:

morpeh [17]3 years ago

5 0

Your answer will be Radio Waves .

That seems to be the only to make sense. Hope that helps u

Blizzard [7]3 years ago

5 0

Examples of Electromagnetic Waves. RADIO waves, Light waves, thermal radiation, X ray, visible light, microwave, infrared, gamma rays etc. are the example of electromagnetic waves. These waves together form the electromagnetic spectrum. (https://physics.tutorvista.com/waves/electromagnetic-waves.html)

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Which one of these statements true about the moderator and the control rods a nuclear power station?

siniylev [52]

Answer: the moderator slows the neutrons down, while the control rods absorb the neutrons.

Explanation:

4 0

3 years ago

A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a sp

cluponka [151]

Answer:

(a) Approximately $0.335\; \rm m$ .

(b) Approximately $1.86\; \rm m\cdot s^{-1}$ .

(c) Approximately $0.707\; \rm m$ .

(d) Approximately $0.228\; \rm m$ .

Explanation:

$v_i$ denotes the velocity of the object in the first diagram right before it came into contact with the spring.
Let $m$ denote the mass of the block.
Let $\mu$ denote the constant of kinetic friction between the object and the surface.
Let $g$ denote the constant of gravitational acceleration.
Let $k$ denote the spring constant of this spring.

Consider the conversion of energy in this object-spring system.

First diagram: Right before the object came into contact with the spring, the object carries kinetic energy $\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2$ .

Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.

Third diagram: After the velocity of the object becomes zero, it has moved a distance of $D$ and compressed the spring by the same distance.

Energy lost to friction: $\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D$ .
Elastic potential energy that the spring has gained: $\displaystyle \frac{1}{2}\,k\, D^2$ .

The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:

$\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

Assume that $g = 9.81\; \rm m \cdot s^{-2}$ . In the equation above, all symbols other than $D$ have known values:

$m =1.10\; \rm kg$ .
$v_i = 2.60\; \rm m \cdot s^{-1}$ .
$\mu = 0.250$ .
$g = 9.81\; \rm m \cdot s^{-2}$ .
$k = 50.0\; \rm N \cdot m^{-1}$ .

Substitute in the known values to obtain an equation for $D$ (where the unit of $D\!$ is $m$ .)

$3.178 = 2.69775\, D + 25\, D^2$ .

$2.69775\, D + 25\, D^2 + 3.178 = 0$ .

Simplify and solve for $D$ . Note that $D > 0$ because the energy lost to friction should be greater than zero.

$D \approx 0.335\; \rm m$ .

The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:

$\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J$ .

As the object moves to the left, part of that energy will be lost to friction:

$(\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J$ .

The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:

$2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J$ .

Calculate the velocity corresponding to that kinetic energy:

$\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}$ .

As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ( $1.91\; \rm J$ ) would be lost to friction.

How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is $\mu \cdot m \cdot g$ .

$\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m$ .

Similar to (a), solving (d) involves another quadratic equation about $D$ .

Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) $1.91\; \rm J$ .

Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.

$\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

$25\, D^2 + 2.69775\, D - 1.90811\approx 0$ .

Again, $D > 0$ because the energy lost to friction is greater than zero.

$D \approx 0.228\; \rm m$ .

7 0

3 years ago

What is the formula of Kinetic Energy

Andre45 [30]

K.E = ½ mv²

Explanation:

The formula of Kinetic Energy

K.E = ½ mv²

Where,

m = mass of object

v = velocity of object

The <u>Standard Unit</u> of kinetic energy is <u>Joule</u>, while the <u>Imperial Unit</u> of kinetic energy is <u>foot-pound</u>

<u>-TheUnknown</u><u>Scientist</u>

8 0

3 years ago

Read 2 more answers

A converging lens brings rays of light together at a focal point. The bending of light rays is the result ofA. A combination of

Tcecarenko [31]

Answer:

.C. Refraction of light passing through the lens.

Explanation:

The convergence or divergence of the rays of light is only due to refraction of light. When a ray of light is made to pass through a lens, it bend's from it's original path. This bending of light from its original path is called Refraction. Whereas in Reflection the ray of light is reflected back into the same medium.

Hence the correct option is C. Refraction of light passing through the lens.

5 0

3 years ago

Q/C An undersea earthquake or a landslide can produce an ocean wave of short duration carrying great energy, called a tsunami. W

iragen [17]

The amplitude $A_{2}$ is 8.307 m

When describing the peak value of a quantity, such as the level of sound waves or the power and voltage in electrical and electronic systems, the term amplitude is employed. A louder sound has a bigger amplitude, and a softer sound has a smaller amplitude.

As we previously discussed in the concept session, the square root of the water's speed is inversely related to the amplitude of its waves.

$A\alpha\frac{1}{\sqrt{v} }$