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3 years ago

why is the meteor crater in arizona still visible when most impact craters created by meteorites hitting earth are not

2 answers:

8 0

Good question. The Arizona impact location is very, very, unique. The crater was the Bigger than Cuba when it first struck. Due to sedimentary drop off its size is slowly decaying. It is currently about the size of 2-3 Carnival cruise ships.

joja [24]3 years ago

4 0

Answer:

The meteor crater in Arizona is still visible due to the sedimentary nature of the materials present, mainly limestone, and the total absence of lava in the zone.

Most of the craters created by other meteorites that have collided with the earth are not visible at present because they were smaller and also impacted in areas where the soil materials were more compact and firm.

Explanation:

It is estimated that the impact produced by the crater 50,000 years ago, and was caused by an nickel-iron meteorite about 50 m long traveling at an approximate speed of 12 km/sec. The impact created a temperature so extraordinarily high that it vaporized or melted all the rocks in the surrounding crust.

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A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves

KIM [24]

Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

Explanation:

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

4 0

3 years ago

11. Which of the following phenomena is taking place when sound waves are reflected from a surface along parallel lines? A. Refr

allsm [11]

The Answer is C. Focusing

4 0

2 years ago

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

$\boxed{B = 0 T}$

3 0

2 years ago

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

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

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

$r'=\frac{1}{4}r$

So the new acceleration is

$a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a$

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

5 0

3 years ago

a 3000 kg and a 7000 kg Mass attract each other with a force of 0.0015 N. What distance separates the two objects (Radius) (Plea

frutty [35]

Answer:

Explanation:

According to law of gravitation;

F = GMm/d²

G is the universal gravitation

M and m are the masses

d is the distance between the masses

d² = GMm/F

d² = 6.67408 × 10-11 *3000*7000/0.0015

d² = 140.15568*10^-5/0.0015

d² = 1.4016*10^-3/0.0015

d² = 1.4016*10^-3/1.5*10^-3

d² = 0.9344*10

d² = 9.344

d = √9.344

d = 3.057m

Hence the distance between the two objects is 3.057m

3 0

3 years ago