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3 years ago

5

How strong does an earthquake have to be to cause a tsunami

1 answer:

Vesnalui [34]3 years ago

8 0

Answer:

An earthquake with the magnitude of more than 7.5 on the Richter scale can cause to a destructive Tsunami.

Explanation:

According to International Tsunami Information center, the cause of the most destructive tsunami is shallow and large earthquake with the epicenter or fault line on the ocean floor or near the ocean floor.
When the tectonic plates (are pieces of earth's crust and uppermost mantle) are collided with each other then the earthquake occurs. This results in the displacement of large area of ocean floor from few km to 1000 km or more. This displacement of ocean floor in the formation of tsunami waves.
Like in the 1960, the tsunami in Chile was resulted due to earthquake with the magnitude of 9.5 on Richter scale with rupture area of over 1000 km. This tsunami not only cause destruction in Chile but also affected the other countries like Japan.

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In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

3 0

2 years ago

In an experiment, you find the mass of a cart to be 250 grams. What is the mass of the cart in kilograms?

gizmo_the_mogwai [7]

It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left

5 0

3 years ago

Read 2 more answers

A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25

solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

$k=\frac {F}{x}$

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

$k=\frac {80}{0.25}=320N/m$

Therefore, the spring constant is equal to 320 N/m

3 0

3 years ago

Which of the following is typically NOT a method scientist use to determine astronomic distances

tangare [24]

I think the answer is A

5 0

3 years ago

In a certain region of space, the electric potential is V(x,y,z)=Axy-Bx^2+Cy , where A,B , and C are positive constants.a) Calcu

laiz [17]

Answer:

a) Eₓ = - A y + 2B x , b) Ey = -Ax –C , c) Ez = 0 , d) The correct answer is 3

Explanation:

The electric field and the electric power are related

E = - dV / ds

a) Let's find the electric field on the x axis

Eₓ = - dV / dx

dV / dx = A y - B 2x

Eₓ = - A y + 2B x

b) calculate the electric field on the y-axis

Ey = - dV / dy

dV / dy = A x + C

Ey = -Ax –C

c) the electric field on the z axis

dv / dz = 0

Ez = 0

.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

0 = = - A y + 2B x

y = 2B / A x

Axis y

0 = -Ax –C

.x = -C / A

We substitute this value in the previous equation

.y = 2B / A (-C / A)

.y = 2 B C / A2

The correct answer is 3

6 0

3 years ago