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2 years ago

What are the 4 rules of seismic waves?

1 answer:

3 0

"The are many different seismic waves, but all of basically of four types:

Compressional or P (for primary)
or S (for secondary)
Love
Rayleigh
An earthquake radiates P and S waves in all directions and the interaction of the P and S waves with Earth's surface and shallow structure produces surface waves."

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How does climate change?

Korvikt [17]

Answer:

Con el cielo del agua

Explanation:

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1 year ago

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A balloon filled with helium has a small hole on the right side of the balloon. How would Pascal explain why the entire balloon

Morgarella [4.7K]

Answer:

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2 years ago

Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin

tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a. $D=A+B+C$

$D=3i-3j+i-4j-2i+5j$

$D=2i-2j$

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude= $\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2$ units

By using the formula $\mid r\mid=\sqrt{x^2+y^2}$

Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

$E=-3i+3j-i+4j-2i+5j$

$E=-6i+12j$

$\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4$ units

Direction of E= $\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}$

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2 years ago

If a coin has a mass of 29.34g and drops from a height of 14.7 meters, what is its loss in gravitational potential energy? Show

Serga [27]

$U= mgh$
$U=((29.34/1000)kg)(9.81 \frac{m}{s^{2} } )(14.7m)=4.321J$

do not forget to change grams to kilograms and if it were in centimeters convert them to meters.

7 0

2 years ago

A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res

VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

$KE = PE$

$\frac{1}{2} mv^2 = mgh$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(2(55.8*10^{-2}))}$

$v = 4.67m/s$

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

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2 years ago