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3 years ago

I need help on 2/3 please asap thx ♥️

Physics

1 answer:

ch4aika [34]3 years ago

5 0

<u>questions 2</u>

F=m

therefore

a=F/m

a=408/68

=6m/s^2.

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Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling

Eva8 [605]

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

$vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2$

Replacing:

$Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9$ $Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9$ $Vf1=-1-2.6=-3.6\text{ m/s}$

Answer: 3.6 m/s west

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1 year ago

A beam of white light enters a prism of crown glass (n = 1.5) from air (n = 1.00). Once inside, the colors in the light disperse

eduard

Answer:

= 2.33

Explanation:

.According to snell's law:

n1sin i = n2sin r ,

where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,

i is angle of incidence,

r is angle of refraction.

Given that,

n1 = 1,

i = 51 degrees,

r = 19.5 degrees. ,

n2= ?

So,

1*sin 51 = n2 sin 19.5

=> n2 = sin51 / sin19.5

= 2.33

7 0

3 years ago

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

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3 years ago

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The answer is allotropes. Hope this helps. Have a great day.

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