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3 years ago

Where does the energy of an earthquake originate?

Physics

2 answers:

viva [34]3 years ago

6 0

A is the correct answer.

expeople1 [14]3 years ago

4 0

A. Movement of tectonic plates... When these plates move they create an earthquake

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A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i

kiruha [24]

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

$\Delta A = 8.00-3.00= 5.00\ A$

We need to calculate the time interval

Using formula of inductor

$V=L\dfrac{\Delta A}{\Delta t}$

$\Delta t =\dfrac{L\Delta A}{V}$

Where, $\Delta A$ = change in current

V = voltage

L = inductance

Put the value into the formula

$\Delta t=\dfrac{1.20\times5.00}{12.00}$

$\Delta t=0.5\ sec$

Hence, The time is 0.5 sec.

5 0

3 years ago

1. What coefficients would balance the following equation?

maksim [4K]

Answer:

D. 4Al + 3O2 → 2Al2O3

Explanation:

Chemical reactions involves the chemical combination of two or more substances called REACTANTS to yield other substances called PRODUCTS. However, in accordance with the LAW OF CONSERVATION OF MASS, the amount of reactants must be equal to that of the products.

To accomplish this, the reaction must be BALANCED. A balanced equation is an equation in which the number of atoms of each element in the reactant side equals the number of atoms in the product side. In this reaction involving Aluminum and Oxygen to give Aluminum oxide as follows:

Al + 02 → Al2O3

A coefficient is used to balance the number of atoms on both sides of the equation as follows:

4Al + 3O2 → 2Al2O3

3 0

3 years ago

Read 2 more answers

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Brums [2.3K]

The car’s velocity at the end of this distance is 18.17 m/s.

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

Final velocity, V = 18.17 m/s

Therefore, the car’s velocity at the end of this distance is 18.17 m/s.

Read more: brainly.com/question/8898885

8 0

2 years ago

Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

solmaris [256]

Answer:

$\lambda = 3.68 *10^{-36} \ m$

$\lambda_p = 1.28*10^{-14} \ m$

Explanation:

From the question we are told that

The mass of the person is $m = 180 \ kg$

The speed of the person is $v = 1 \ m/s$

The energy of the proton is $E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{h}{m * v }$

Here h is the Planck constant with the value

$h = 6.62607015 * 10^{-34} J \cdot s$

$\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }$

=> $\lambda = 3.68 *10^{-36} \ m$

Generally the energy of the proton is mathematically represented as

$E_p = \frac{1}{2} * m_p * v^2_p$

Here $m_p$ is the mass of proton with value $m_p = 1.67 *10^{-27} \ kg$

=> $8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2$

=> $v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }$

=> $v = 3.09529 *10^{7} \ m/s$

$\lambda_p = \frac{h}{m_p * v_p }$

so $\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }$

=> $\lambda_p = 1.28*10^{-14} \ m$

5 0

3 years ago

On his way off to college, Russell drags his suitcase 19 m from the door of his house to the car at a constant speed with a hori

Mashcka [7]

Answer:

The work done on the suitcase is, W = 1691 J

Explanation:

Given data,

The force on the suitcase is, F = 89 N

The distance Russell dragged the suitcase, S = 19 m

The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction

The work done on the suitcase by Russell is given by the formula

W = F · S

Substituting the given values,

W = 89 N x 19 m

W = 1691 J

Hence, the work done on the suitcase is, W = 1691 J

8 0

4 years ago