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3 years ago

At this rate, how long would it take for two continents 3500 kilometers apart to collide?

Physics

1 answer:

meriva3 years ago

5 0

Given:

3,500 kilometers

Find:

Years for two continents to collide = ?

Solution:

We know that typical motions of one plate relative to another are 1 centimeter per year.

So first, we convert 3,500 km to cm.

The solution would be like this for this specific problem:

1 km = 100,000 cm

3,500 km x 100,000 = 350,000,000 cm

Since we know that 1 cm = 1 year, then that means 350,000,000 cm is equivalent to 350,000,000 years.

Therefore, it would take 350 million years for two continents that are 3500 kilometers apart to collide.

To add, a phenomenon of the plate tectonics of Earth that occurs at convergent boundaries is called the continental collision.

You might be interested in

What two kinds of crust are involved in a subduction zone

Anarel [89]

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

5 0

3 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

7 0

2 years ago

A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3

PtichkaEL [24]

Answer:2.47

Explanation:

So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!

7 0

2 years ago

Un móvil se encuentra en la posición inicial x0 = 22 m, y se mueve con velocidad 5 m/s. Calcula la posición en la que se encuent

Brilliant_brown [7]

Answer:

La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.

Explanation:

El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante.

La distancia recorrida, x , por un móvil que tiene un MRU con un velocidad v durante el intervalo de tiempo t es:

x= x0 + v*t

donde x0 es la posición inicial.

En este caso:

x0= 22 m
v= 5 m/s
t= 30 s

Reemplazando:

x= 22 m + 5 m/s* 30 s

Resolviendo:

x= 22 m + 150 m

x= 172 m

La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.

8 0

2 years ago

A man at point A directs his rowboat due north toward point B, straight across a river of width 100 m. The river current is due

prohojiy [21]

Answer:

1.35208 m/s

Explanation:

Speed of the boat = 0.75 m/s

Distance between the shores = 100 m

Time = Distance / Speed

$Time=\frac{100}{0.75}=133.33\ s$

Time taken by the boat to get across is 133.33 seconds

Point C is 150 m from B

Speed = Distance / Time

$Speed=\frac{150}{\frac{100}{0.75}}=1.125\ m/s$

Velocity of the water is 1.125 m/s

From Pythagoras theorem

$c=\sqrt{0.75^2+1.125^2}\\\Rightarrow c=1.35208\ m/s$

So, the man's velocity relative to the shore is 1.35208 m/s

3 0

3 years ago