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ikadub [295]
3 years ago
13

1. What is the relationship between wavelength and the penetrating power of an electromagnetic wave?

Physics
1 answer:
Andreyy893 years ago
6 0

#1

As we know that energy of electromagnetic wave is given by

E = \frac{hc}{\lambda}

so here we know that penetrating power will directly depends on its energy and energy inversely depends on wavelength

So here we can say correct answer will be

C) The penetrating power decreases as the wavelength increases.

#2

Speed of sound is maximum in solids and minimum in gas

so here as ice melts into water the speed of sound must have to decrease

so correct answer will be

D) The speed of sound would decrease because sound travels faster through solids than liquids.

#3

mechanical waves required medium to travel while non mechanical waves do not require any medium to travel

so here correct answer will be

A) sound

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Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i
Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

L = L_i_n_i + L_3_s

L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt

L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s

Moment if inertia is

I = 2ml^2

I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2

Angular speed

ω = L/I

= 38 / 5.75\\\\=6.61

The speed of each ball is

V = ωL

= 6.61\times0.5\\\\= 3.31m/s

7 0
3 years ago
Select the correct answer.
spin [16.1K]

Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

7 0
3 years ago
0.0084x91 ...................
Drupady [299]

Answer:

0.0084×91=0.7644

Explanation:

hope this helps

8 0
3 years ago
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A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.
melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

V_{f}^{2} = V_{0}^{2} + 2ad

<u>Where</u>:

V_{f}: is the final speed = 8.89 m/s

V_{0}: is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m    

a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}

Therefore, the magnitude of her acceleration is 3.09 m/s².              

b) The component of her acceleration that is parallel to the ground is given by:

a_{x} = a*cos(\theta)

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0
3 years ago
When positively charged particles were radiated onto a gold atom, a few of the particles bounced back. Which of the following ca
ratelena [41]

Positively charged protons in the nucleus of the gold atom .... rutherford scattering ???


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3 years ago
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