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3 years ago

An earthquake’s epicenter is _____.

Physics

2 answers:

Evgen [1.6K]3 years ago

4 0

The point in which it originates.

Anastasy [175]3 years ago

4 0

Hi, hope you’re having a good day.

An earthquake epicenter is the point on the earth's surface above the hypocenter (or focus), point in the crust where a seismic rupture begins.

Idk, you could search more on the internet but if you need help understanding more, you can just comment it and I’ll try to reply ASAP.

Anyways, thanks for reading, have an amazing day, and stay happy. ❤️❤️

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A fan blade is rotating with a constant angular acceleration of 11.5 rad/s2. At what point on the blade, as measured from the ax

vichka [17]

Answer:

Alpha = ω^2 R where R is radius of blade

g = w^2 r where r is distance from center

ω^2 R = 11.5 ω^2 r

R / r = 11.5 / 9.8 = 1.17

Or r = .852 R

Since the angular acceleration depends on both R and ω it seems that one can only get r as it depends on R

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2 years ago

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3 years ago

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You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m.

Gwar [14]

Answer:

0.95 seconds

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s² (downward positive, upward negative)

Time taken by the ball to reach the maximum height

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-15}{-9.81}\\\Rightarrow t=1.52\ s$

Maximum height

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-15^2}{2\times -9.81}\\\Rightarrow s=11.47\ m$

Distance between maximum height of the ball and the branch is 11.47-7 = 4.46 m

So, the distance that will be covered on the way down is 4.46 m

Now

u = 0

s = 4.46

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.46=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4.46\times 2}{9.81}}\\\Rightarrow t=0.95\ s$

Time taken by the ball from the maximum height to the tree branch is 0.95 seconds.

Total time taken from the moment the ball is thrown to reach the tree branch is 1.52+0.95 = 2.47 seconds

7 0

3 years ago

. (Serway 9th ed., 7-33) A 0.600-kg particle has a speed of 2.00 m/s at point A and a kinetic energy of 7.50 J at point B. What

k0ka [10]

Answer:

a) KA = 1.2 J

b) vB = 5.00 m/s

c) W = 6.30 J

Explanation:

m = 0.600 kg

vA = 2.00 m/s

KB = 7.50 J

a) KA = ?

b) vB = ?

c) W = ?

We can apply the folowing equations

K = 0.5*m*v²

and

W = ΔK = KB- KA

then

a) KA = 0.5*m*vA² = 0.5*(0.600 kg)*(2.00 m/s)²

⇒ KA = 1.2 J

b) KB = 0.5*m*vB² ⇒ vB = √(2*KB / m)

⇒ vB = √(2*7.50 J / 0.600 kg)

⇒ vB = 5.00 m/s

c) W = ΔK = KB- KA = (7.50 J) - (1.2 J)

⇒ W = 6.30 J

5 0

4 years ago