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3 years ago

11

Compare these two waves

1 answer:

otez555 [7]3 years ago

4 0

None of the choices in the picture is correct.

The blue wave has a higher pitch than the orange one, (because its waves are shorter so there are more of them in the same length of time).

AND the blue wave is also louder than the orange one, (because the amplitude of its waves is slightly greater).

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An electric heater is operated by applying a potential difference of 78.0 V across a nichrome wire of total resistance 7.00 Ω. a

aleksley [76]

PART A)

Here we know that

potential difference across the wire is

$V = 78 volts$

resistance of wire is

$R = 7 ohm$

now by ohm's law

$V = iR$

$78 = i(7 ohm)$

$i = 11.14 A$

Part b)

Power rating is defined as rate of electrical energy

it is defined as

$P = iV$

now we have

$P = 11.14 ( 78)$

$P = 869.1 Watt$

8 0

3 years ago

Obligation of reasearchers to review true or fasle

MrMuchimi

I think if I was a researcher I would review my work or the sources I use, True

5 0

4 years ago

NEED HELP ASAP PLEASE

Zarrin [17]

Answer:

D. 3.0 m/s

Explanation:

because I did this in my class

8 0

3 years ago

Which charateristic of the observant function?

valentina_108 [34]

Answer:

All people with the observant trait are often a steadying force that always wants to get things done. Their energy is very elegant in the sense of working on real things in real time.

Explanation:

Hope I helped

8 0

2 years ago

In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with t

Mamont248 [21]

Answer:

a) $h_{max}=14536.16 m$

b) $h = 15687.9 m$

c) $PD=7.62\%$ The estimate is low.

Explanation:

a) Using the energy conservation we have:

$E_{initial}=E_{final}$

we have kinetic energy intially and gravitational potential energy at the maximum height.

$\frac{1}{2}mv^{2}=mgh_{max}$

$h_{max}=\frac{v^{2}}{2g}$

$h_{max}=\frac{43^{2}}{2*0.0636}$

$h_{max}=14536.16 m$

b) We can use the equation of the gravitational force

$F=G\frac{mM}{R^{2}}$ (1)

We have that:

$F = ma$ (2)

at the surface G will be:

$G=\frac{gR^{2}}{M}$

Now the equation of an object at a distance x from the surface.

is:

$F=\frac{mgR^{2}}{(R+x)^{2}}$

$m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}$

Using that dv/dt is vdx/dt and integrating in both sides we have:

$v_{0}=\sqrt{\frac{2gRh}{R+h}}$

$h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}$

$h=15687.9$

c) The difference is:

So the percent difference will be:

$PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%$

$PD=7.62\%$

The estimate is low.

I hope it helps you!

7 0

3 years ago