All categories

2 years ago

What landforms form at convergent boundaries where two oceanic plates collide?

2 answers:

6 0

They are formed when two plates collide, either crumpling up and forming mountains or pushing one of the plates under the other and back into the mantle to melt. Convergent boundaries form strong earthquakes, as well as volcanic mountains or islands, when the sinking oceanic plate melts.

Bad White [126]2 years ago

4 0

It forms mountains and sometimes islands.

You might be interested in

The constant pressure molar heat capacity, C_{p,m}C p,m , of nitrogen gas, N_2N 2 , is 29.125\text{ J K}^{-1}\text{ mol}^{-1

antoniya [11.8K]

Answer:

Explanation:

Constant pressure molar heat capacity Cp = 29.125 J /K.mol

If Cv be constant volume molar heat capacity

Cp - Cv = R

Cv = Cp - R

= 29.125 - 8.314 J

= 20.811 J

change in internal energy = n x Cv x Δ T

n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature

Putting the values

= 20 x 20.811 x 15

= 6243.3 J.

3 0

2 years ago

¿Alguien me puede ayudar? Problema: Un niño le pide gastada a su papá y éste le contesta que le dará los $120 que tiene en su bo

VLD [36.1K]

Answer: there are 15 coins of $2 and 18 coins of $5

Explanation:

I will answer in English.

X is the number of $5 coins.

Y is the number of $2 coins.

We have the system of equations:

Y + X = 33

Y*2 + X*5 = 120

first, we must isolate one of the variables in one of the equations and then replace it in the other equation, let's isolate Y in the first equation:

Y = 33 - X.

Then we can replace it in the other equation:

(33 - X)*2 + X*5 = 120

66 - X*2 + X*5 = 120

X*3 = 54

X = 54/3 = 18

and using the equation for Y.

Y = 33 - X = 33 - 18 = 15

So there are 15 coins of $2 and 18 coins of $5

3 0

2 years ago

How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov

sveta [45]

Answer:

x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This is a projectile launch exercise, in this case we will write the equations for the x and y axes

Let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / v₀

cos θ = v₀ₓ / v₀

v_{y} = v_{oy} sin θ

v₀ₓ = vo cos θ

now let's write the equations of motion

X axis

x = v₀ₓ t

x = v₀ cos θ t

vₓ = v₀ cos θ

Y axis

y = y₀ + $v_{oy}$ t - ½ g t2

y = y₀ + v₀ sin θ t - ½ g t2

v_{y} = v₀ - g t

v_{y} = v₀ sin θ - gt

$v_{y}^{2}$ = v_{oy}^2 sin² θ - 2 g y

As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components

7 0

2 years ago

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

3 0

2 years ago

Read 2 more answers

A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha

Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

$PQ = \sqrt{4^{2}+3^{2}}=5 m$

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

$E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C$

Electric field at P due to the charge q is

$E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C$

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

$E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C$

3 0

2 years ago