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2 years ago

What is the atomic bomb equivalent to a 7.0 earthquake

Physics

1 answer:

Svetllana [295]2 years ago

5 0

Answer:

199,000 tons.

Explanation:

A 7.0 earthquake has an equivalence of 199,000 tons of TNT.

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Which of the following statements are true? Check all that apply.

lidiya [134]

We have 6 levels and we will verify which of them is true.

A ) Electric charge is conserved

According to the conservation of charge, an electric charge can neither be created nor be destroyed.

Therefore this statement is TRUE.

B ) A positive charge and negative charge attract each other

According to the properties of charges, the opposite charges attract each other and similar charges repel each other

Therefore this statement is TRUE

C) Two positive charges attract each other.

Using the previous law, we can observe that this statement is false, because equal charges repel each other.

Therefore this statement is FALSE.

D ) Two negative charges repel each other.

Using the same principle as the two previous statements, two equal charges repel each other.

Therefore this statement is TRUE

E) Electric charge is quantized.

According to the quantization of charge, charges can take values only in integral multiples of a fundamental unit of charge.

Therefore this statement is TRUE.

F) A neutral object contains no charge.

A neutral object contains equal amount of negative and positive charge. Thus, in totality a neutral object contains no charge.

Therefore this statement is TRUE.

8 0

2 years ago

According to the three laws of planetary motion, planetary orbits are in the shape of a/an

Talja [164]

According to the three laws of planetary motion, planetary orbits are in the shape of an "Ellipse"

In short, Your Answer would be Option B

Hope this helps!

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3 years ago

Why is density used to determine the identity of matter?

dedylja [7]

a substance's density is the same at a certain pressure and temperature, and the density of one substance is usually different than another substance.

4 0

2 years ago

iris [78.8K]

Answer:

the answer is a time your welcome

3 0

2 years ago

Read 2 more answers

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$

5 0

3 years ago