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3 years ago

Green light travels by what type of wave? (2 points) a Conductive b Electronic c Conducive d Electromagnetic

Physics

2 answers:

antiseptic1488 [7]3 years ago

8 0

Answer: d. Electromagnetic wave

Explanation:

Light is an electromagnetic wave, no matter is wavelength or color.

In this sense, the main characteristic of this type of wave is that it can propagate trhough vacumm, it is not necessary a medium. In addition, electromagnetic waves are transversal waves, this means the electric field oscillates in all normal directions to the direction of wave propagation.

Rom4ik [11]3 years ago

4 0

Answer:

d) Electromagnetic waves

Explanation:

As we know that electromagnetic waves are of 7 types as per their energy range

these are as following

1) Gamma Rays

2) X Rays

3) Ultraviolet rays

4) Visible light (VIBGYOR all colors)

5) Infrared waves

6) microwaves

7) Radio waves

Since all type of visible light exist in the range of electromagnetic waves

so we can say that Green light travels as per nature of electromagnetic waves

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The time, t, required to drive a fixed distance varies inversely as the speed, r. It takes 3 hr at a speed of 14 km/h to driv

kenny6666 [7]

Answer:

time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes

Explanation:

Given:

Time is inversely proportional to the speed

mathematically,

t ∝ (1/r)

let the proportionality constant be 'k'

thus,

t = k/r

therefore, for case 1

time = 3 hr

speed = 14 km/hr

3 = k/14

also,

for case 2

let the time be = t

r = 23 km/h

thus,

we have

t = k/23

on dividing equation 2 by 1

we get

$\frac{t}{3}=\frac{k/23}{k/14}$

$t=\frac{14\times3}{23}$

t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)

4 0

3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :