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erma4kov [3.2K]
3 years ago
5

Which of these waves moves the fastest? a. sound b. radio c. ocean d. seismic

Physics
2 answers:
Darya [45]3 years ago
5 0
The fastest wave is Radio wave.

Answer is B.

Hope it helps!
NARA [144]3 years ago
3 0
First of all, the question is vague as you haven't mentioned the medium of propagation of the wave, which is extremely crucial.

For example light travels at <span>299,792</span> km/second in vacuum, but in certain semiconductors, it travels as slow as 9 km/second. Sound, ocean and seismic waves don't exist in vacuum at all.

If you mean the maximum possible speed any of these options can attain in any medium of choice for the different options, then the answer would be 
b. radio waves, which travel exactly at the speed of light in vacuum (299,792 km/second) and with an almost similar(slightly less) speed in air. (Radio waves are nothing but electromagnetic waves with low frequency)
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What is in the blank?
Vika [28.1K]
This is in the thermosphere which is at an altitude of 85-520km
8 0
3 years ago
A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0
3 years ago
A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

(c) Us ethe conservation of energy

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0
3 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

#SPJ1

5 0
2 years ago
A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
2 years ago
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