Ask question

Ask question

All categories

3 years ago

11

What are the three key points to seismic waves

1 answer:

leonid [27]3 years ago

6 0

Earthquakes produce three types of seismic waves: primary waves, secondary waves, and surface waves. Each type moves through materials differently. In addition, the waves can reflect, or bounce, off boundaries between different layers.

three key points:

1. they all change the earth by shifting plates in different ways.

2. these waves lead to our plates form today, which are always changing.

3. even the smallest earthquakes make the slightest changes to our ground. and many happen in just a day! we just never feel it.

You might be interested in

Most of the stars in the Milky Way will end their lives as

umka21 [38]

<span> </span>Most of the stars in the Milky Way will end their lives as white dwarfs

4 0

2 years ago

Read 2 more answers

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

$M = 5.98 \times 10^{24} kg$

$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

$R = 6.37 \times 10^6 m$

now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

$v = 7934.2 m/s$

5 0

3 years ago

The escape velocity at the surface of Earth is approximately 11 km/s. What is the mass, in units of ME (the mass of the Earth),

SVETLANKA909090 [29]

Answer:11 km/s

Explanation:

Given

Escape velocity at the surface of earth is 11 km/s

Escape velocity is given by

$V_e=\sqrt{\frac{2GM}{R}}$

Escape velocity at the surface of earth

$11=\sqrt{\frac{2GM}{R}}$ --------------------1

If Escape velocity is three times and the radius is also the three times

$V_e_2=\sqrt{\frac{2G(3M)}{3R}}$

$V_e_2=\sqrt{\frac{2GM}{R}}=V_e$

i.e. $V_e_2=11 km/s$

5 0

2 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

2 years ago

Heat can be transferred in several ways. Which of the following is an example of conduction? (DOK 2) A The sun is shining on a m

forsale [732]

Answer:

Explanation:

G yvyvybybybububunununu

8 0

3 years ago