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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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jasenka [17]

Answer:

B. 0.1 meters/second/second in the same direction of travel.

Explanation:

Acceleration is the rate of change of velocity. Acceleration is a vector quantity.

a=Δv/Δt

=(v₂-v₁)/(t₂-t₁)

v₁=4 m/s

v₂=6 m/s

t₁=0 s

t₂=20 s

a=(6m/s-4m/s)/(20s-0)

= 0.1 m/s² in the same direction of travel.

Therefore acceleration =0.1 meters/second/second in the same direction of travel.

7 0
3 years ago
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If a biker starts at rest and accelerates to 15 m/s in 7.5 seconds, what is his<br> acceleration? *
sergeinik [125]

Answer:

2 m/s^2

Explanation:

from the question

v=15 m/s

t=7.5

a=?

from the first equation of motion

v=u+at

where,

v=final velocity

u=initial velocity

a=acceleration

t=time

from the question (u) will be zero because the body started at rest

v=u+at

15=(0)+a×7.5

15=7.5a

a=15/7.5

a=2 m/s^2

6 0
3 years ago
A 20 cm
musickatia [10]

Answer:

4.8.

Explanation

7 0
3 years ago
If the ladybug has a mass of 6.0 kg and is at a distance r = 1 m, find her A) velocity, B) centripetal acceleration, and C) cent
Dominik [7]

a) 3.14 m/s

b) 9.9 m/s^2

c) 59.4 N

Explanation:

a)

The relationship between angular velocity and linear velocity for an object in circular motion is given by:

v=\omega r

where

\omega is the angular velocity

v is the linear velocity

r is the distance between the object and the axis of rotation

For the ladybug in this problem, we have:

r = 1 m (distance)

\omega=3.14 rad/s (angular velocity)

So, its linear velocity is

v=(3.14)(1)=3.14 m/s

b)

The centripetal acceleration of an object in circular motion is given by the equation

a=\frac{v^2}{r}

where

v is the linear velocity

r is the distance between the object and the axis of rotation

The centripetal acceleration represents the radial acceleration of the object.

For the ladybug in this problem, we have:

v = 3.14 m/s (linear velocity)

r = 1 m (distance)

So, the centripetal acceleration is

a=\frac{3.14^2}{1}=9.9 m/s^2

c)

The centripetal force for an object in circular motion is given by

F=ma

where

m is the mass of the object

a is the centripetal acceleration of the object

The centripetal force is the radial inward force that keeps the object in circular motion.

For the ladybug in this problem we have

m = 6.0 kg (mass)

a=9.9 m/s^2 (centripetal acceleration)

So, the centripetal force is

F=(6.0)(9.9)=59.4 N

6 0
3 years ago
Evaluate cosπ/11+cos3π/11+cos5π/11+cos7π/11+cos9π/11
alexandr402 [8]
Using a calculator, the answer will be 1/2 as a result of all the / and use of opposite πs that form 18 making a whole circle.
4 0
3 years ago
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