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3 years ago

What would happen if an Asteroid hit earth?

Physics

2 answers:

svp [43]3 years ago

4 0

Answer:

There would be tremendous earthquakes and tsunamis, followed by massive volcanism around the impact zone. The ozone layer would be destroyed. The oceans would turn acidic. The Sun would be blotted out, probably for decades.

Explanation:

Hope this helps!

Mars2501 [29]3 years ago

4 0

Answer: Well it depends on the size. If the asteroid was the size of a basketball, it would most probably turn to dust by the time it reaches the surface. But, if it was the size of Australia, then most probably everyone would die.

Explanation:

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A 1180kg car is moving at a speed of 85.5 km/h. Find the force needed to bring the car to rest in a distance of 300m

yarga [219]

Answer:

1110 N

Explanation:

First, find the acceleration.

Given:

Δx = 300 m

v₀ = 85.5 km/h = 23.75 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (23.75 m/s)² + 2a (300 m)

a = -0.94 m/s²

Find the force:

F = ma

F = (1180 kg) (-0.94 m/s²)

F = -1110 N

The magnitude of the force is 1110 N.

3 0

3 years ago

when a constant force acts upon an object the acceleration of the object varies inversely with its mass. when a certain constant

solong [7]

Explanation:

When a constant force acts upon an object the acceleration of the object varies inversely with its mass.

$a\propto \dfrac{1}{m}$

$\dfrac{a_1}{a_2}=\dfrac{m_2}{m_1}$

If m₁ = 21 kg, a₁ = 3 m/s², m₂ = 9 kg

We need to find a₂

So,

$a_2=\dfrac{m_1a_1}{m_2}\\\\a_2=\dfrac{21\times 3}{9}\\\\a_2=7\ m/s^2$

So, if mass is 9 kg, its acceleration is 7 m/s².

8 0

3 years ago

A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.

tankabanditka [31]

Answer:

$v_o=40.14\ m/s$

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

The horizontal speed is constant and equal to the initial speed $v_o$
The vertical speed is zero at launch time, but increases as the object starts to fall
The height of the object gradually decreases until it hits the ground
The horizontal distance where the object lands is called the range

We have the following formulas

$\displaystyle v_x=v_o$

$\displaystyle x=v_o.t$

$\displaystyle v_y=g.t$

$\displaystyle y=\frac{gt^2}{2}$

Where $v_o$ is the initial horizontal speed, $v_y$ is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

$\displaystyle y=\frac{gt^2}{2}$

Rearranging and solving for t

$\displaystyle 2y=gt^2$

$\displaystyle t^2=\frac{2\ y}{g}$

$\displaystyle t=\sqrt{\frac{2\ y}{g}}$

We then replace this value in

$\displaystyle x=v_o.t$

To get

$\displaystyle v_o=\frac{x}{t}$

$\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}$

$\displaystyle v_o=\sqrt{\frac{g}{2y}}.x$

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing $v_o$

$\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6$

The launch velocity is

$\boxed{v_o=40.14\ m/s}$

7 0

3 years ago

What distance separates Earth and the Sun? A 300,000 km per second B Eight light-minutes C 149 million light-seconds D One light

lina2011 [118]

Answer:

B Eight light-minutes

Explanation:

In the case when the distance separated earth and the sun so here we orbit the sun for a 150 million km distance and the light moves would be 300,000 kilometers per second

Now divide this

= 150 million ÷ 300,000 kilometers per second

= 500 seconds

This 500 seconds represent 8 minutes and 20 seconds

Hence, option B is correct

5 0

3 years ago

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To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

Romashka-Z-Leto [24]

I think the answer is A.

7 0

3 years ago

Read 2 more answers