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3 years ago

How does the time difference between the first p-wave and the first s-wave relate to the distance to the epicenter?

1 answer:

6 0

S-waves do, the seismograph will detect P-waves arriving first, and S-waves will follow. The time difference, as recorded on a clock, between when the P-waves and S-waves ... earthquake waves speed up with increasing distance, and the lag time graph

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(a) In the baggage claim area of an airport, a particular baggage carousel is shaped like a section of a large cone, steadily ro

Tom [10]

Answer:

Part a)

$F_f = 107.8 N$

Part b)

$\mu = 0.415$

Explanation:

Part a)

Time period of one revolution is given as

T = 42 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{42}$

$\omega = 0.15 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.15^2)(7.46)cos20.5 = F_f$

$F_f = 107.8 N$

Part b)

New Time period of one revolution is given as

T = 30 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{30}$

$\omega = 0.21 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.21^2)(7.94)cos20.5 = F_f$

$F_f = 112.9 N$

Also we know that in perpendicular direction also force is balanced

$F_n + m\omega^2 r sin\theta = mgcos\theta$

$F_n = mgcos\theta - m\omega^2 r sin\theta$

$F_n = 30(9.81)cos20.5 - 30(0.21)^2(7.94)sin20.5$

$F_n = 272 N$

now for friction coefficient we will have

$F_f = \mu F_n$

$112.9 = \mu 272$

$\mu = 0.415$

8 0

3 years ago

Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity af

nordsb [41]

Answer:

Explanation:

As the final Kinetic energy is zero or less than initial kinetic energy, the collision must be inelastic.

In Inelastic collision both the bodies must stick together as final velocity is zero for both the bodies.

To conserve the momentum, momentum associated before the collision of first must be equal and opposite to the momentum associated with the second ball.

i.e.

$m_1v_1=m_2v_2$

8 0

3 years ago

Which statement accurately describes the relationship between air pressure, air density, or altitude?

8_murik_8 [283]

Answer:

The answer is: dense air exerts more pressure than less dense air.

Pressure is due to the weight of the above column of the substance which is a force resulting from its weight that is the product of its mass and gravitational pull. The denser the air and the grater the column, the larger its mass and hence the more the pressure it exerts.

I Hope It's Helpful

Hint The Brainliest :)

Explanation:

8 0

4 years ago

Read 2 more answers

True or false. More people live in cities then they do in the country

iVinArrow [24]

Answer:

Explanation:

That's my answer hope it helped

8 0

3 years ago

What force would be required to accelerate a 1,100 kg car to 0.5 m/s2?

Klio2033 [76]

Force =mass x acceleration =1100 x0.5 =550

4 0

3 years ago