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eimsori [14]
4 years ago
12

Meteor Infrasound A meteor that explodes in the atmosphere creates infrasound waves that can travel multiple times around the gl

obe. One such meteor produced infrasound waves with a wavelength of 6.6 km that propagated 8810 km in 8.67 h. Part A What was the frequency of these waves?
Physics
1 answer:
ladessa [460]4 years ago
8 0

Answer:

The frequency of these waves is 4.27\times10^{-2}\ Hz

Explanation:

Given that,

Wavelength = 6.6 km

Distance = 8810 km

Time t = 8.67 hr

We need to calculate the velocity of sound

Using formula of velocity

v = \dfrac{D}{T}

Where, D = distance

T = time

Put the value into the formula

v =\dfrac{8810}{8.67}

v=1016\ km/hr

We need to calculate the frequency

Using formula of frequency

v=n\lambda

n=\dfrac{v}{\lambda}

Put the value into the formula

n=\dfrac{1016}{6.6}

n=153.93\ hr

n=\dfrac{153.93}{60\times60}

n=0.0427\ Hz

n=4.27\times10^{-2}\ Hz

Hence, The frequency of these waves is 4.27\times10^{-2}\ Hz

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Answer:

B

Explanation:

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3 years ago
6. An object moves along the x-axis. The graph shows its position x as a function of time t. Find
raketka [301]

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Explanation:

9-3/5-1= 6/4= 3/2= 1.5

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3 years ago
According to Coulomb's law, what happens to the attraction of two oppositely charged objects as their distance of separation inc
NARA [144]

Answer:

Option B. Decreases

Explanation:

Coulomb's law states that:

F = Kq₁q₂ / r²

Where:

F => is the force of attraction between two charges

K => is the electrical constant.

q₁ and q₂ => are the two charges

r => is the distance apart.

From the formula:

F = Kq₁q₂ / r²

The force of attraction (F) is inversely proportional to the square of their separating distance (r).

This implies that as the distance between them increase, the force of attraction between the two charges will decrease and as the distance between two charges decrease, the force of attraction between them will increase.

Considering the question given above and the illustration given above, the force of attraction will decrease as their distance of separation increases.

Option B gives the right answer to the question.

7 0
3 years ago
Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph .
Serggg [28]

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h

Next, you calculate the difference between both times t1 and t2:

\Delta t=t_1-t_2=2.30h-1.875h=0.425h

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

3 0
3 years ago
The sine of the incident angle is 0.217; the sine of the refracted angle is 0.173. calculate the index of refraction.
notka56 [123]
This can be solve using snell's law. snell's law equation is :

N1 / N2 = sin a2 / sin a1
where N1 is the index of  refraction of the air which is equal to 1
N2 is the index of refraction of the medium
a2 is the angle of refraction
a1 is the incident angle

subsitute the given values
1 / N2 = 0.173 / 0.217
N2 = 1 ( 0.217 / 0.173)
N2 = 1.25 is the index of refraction
5 0
4 years ago
Read 2 more answers
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