The process by which metamorphic rock changes to igneous rock begins with?

A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the ground. How far, in meters, from the deck doe

Vesna [10]

Answer:

2.605m

Explanation:

Using the formula for calculating Range (distance travelled in horizontal direction)

Range R = U√2H/g

U is the speed = 4.8m/s

H is the maximum height = ?

g is the acc due to gravity = 9.8m/s²

R = 3.5m

Substitute into the formula and get H

3.5 = 4.8√2H/9.8

3.5/4.8 = √2H/9.8

0.7292 = √2H/9.8

square both sides

0.7292² = 2H/9.8

2H = 0.7292² * 9.8

2H = 5.21

H = 5.21/2

H = 2.605m

Hence the height of the ball from the ground is 2.605m

7 0

2 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

2 years ago

If earth had no atmosphere , would a falling object ever reach terminal velocity?

Bas_tet [7]

The right answer is no.

6 0

2 years ago

A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the averag

andrey2020 [161]

Answer:

The small car and the truck experience the same average force.

Explanation:

Here we need to remember two of Newton's laws.

The second one says that:

F = m*a

force equals mass times acceleration.

And the third one says that;

"If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A"

From the third law, if the car experiences a force F due to the impact with the truck, then the truck experiences the same force F due to the impact.

But this seems odd, because we would expect to see the car being more affected by the impact, right?

Well, this is explained by the second law.

Suppose that the mass of the car is m, and the mass of the truck is M.

such that M > m

Then for the small car we have:

F = m*a

And for the truck:

F = M*a'

Because the force is the same for both of them, we can write:

m*a = M*a'

a = (M/m)*a'

because M > m, then M/m > 1.

This means that the acceleration that the car experiences is larger than the acceleration for the truck, and this is why we would see that the car seems more affected by the impact, regardless of the fact that both vehicles experience the same force in the impact.

6 0

2 years ago

I know the enthalpy of a reaction is 23kj/mol. Initially the reaction is taking place at 273 k. To what temperature do i need to

Vladimir79 [104]

Answer:

293k

Explanation:

In this question, we are asked to calculate the temperature to which the reaction must be heated to double the equilibrium constant.

To find this value, we will need to use the Van’t Hoff equation.

Please check attachment for complete solution

7 0

3 years ago