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Marysya12 [62]
3 years ago
12

The velocity of the transverse waves produced by an earthquake is 5.09 km/s, while that of the longitudinal waves is 8.5512 km/s

. A seismograph records the arrival of the transverse waves 64.9 s after that of the longitudinal waves. How far away was the earthquake
Physics
1 answer:
mylen [45]3 years ago
8 0

Answer:

Explanation:

Velocity of transverse wave Vt = 5.09 km/s

Velocity of longitudinal wave Vl = 8.5512 km/s

Let earthquake occurred d km  away .

Time taken by transverse wave to travel d distance

= d / 5.09 s

Time taken by longitudinal  wave to travel d distance

= d / 8.5512 s

According to question

\frac{d}{5.09} -\frac{d}{8.5512} = 64.9

.19646 d - .11694 d = 64.9

.07952 d = 64.9

d = 816.15 km .

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Nataly [62]

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

4 0
2 years ago
Sunglasses that reduce glare take advantage of which kind of wave
marysya [2.9K]

Answer:

It should be option B polarization

6 0
3 years ago
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A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur
daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

H = \frac{nI}{l}

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

n = \frac{Hl}{I}

The magnetic field is again given by,

H = \frac{B}{\mu_t}

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

n = \dfrac{\frac{B}{\mu_t}l}{I}

Replacing with our values we have that,

n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}

n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}

n = 27.5 \approx 28

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4 0
3 years ago
What is the relationship between acceleration and time?
Harlamova29_29 [7]

Answer:

The relationship between acceleration and time relates to the velocity and how it changes throughout the movement of an object.

8 0
3 years ago
The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potentia
MAVERICK [17]

Answer:

Energy Lost for group A's car = 0.687 J

Energy Lost for group B's car = 0.55 J

Explanation:

The exact question is as follows :

Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.

To find - How much energy is lost due to heat for group A's car ?

              How much for Group B's car ?

Solution -

We know that,

GPE = 1 Joule (Potential Energy)

Now,

For Group A -

Energy Lost = GPE - KE

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So,

Energy Lost for group A's car = 0.687 J

Now,

For Group B -

Energy Lost = GPE - KE

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So,

Energy Lost for group B's car = 0.55 J

8 0
3 years ago
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