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2 years ago

The waves produced on the earth’s surface is called

Physics

2 answers:

Maslowich2 years ago

6 0

Answer:

seismic wave

Explanation:

These waves are produced because of plate tectonics
These helps are useful in measurements of earthquake.
Used in seismograph.

kotykmax [81]2 years ago

6 0

Answer:

Option A. Seismic wave

Explanation:

The waves produced on the earth's surface during earthquakes are called seismic waves

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Unit Test Review

PolarNik [594]

Answer:

option B...

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2 years ago

Answer and I will give you brainiliest <br><br><br>Please heeeelp

Lostsunrise [7]

Answer:

T = 4.905[N]

Explanation:

In order to solve this problem we must perform a sum of forces on the vertical axis.

∑Fy = 0

We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.

$T-m*g=0\\T=0.5*9.81\\T=4.905[N]$

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3 years ago

Which statement is most likely correct?

DENIUS [597]

Answer:

the correct answer is B

8 0

3 years ago

Read 2 more answers

A pyramid was built with approximately 2.3 million stone blocks, each weighing 2.4 tons (1 ton = 2,000 lbm). Find the mass of th

pogonyaev

Total weight in tons

$\\ \sf\longmapsto 2.3\times 10^6\times 2.4$

$\\ \sf\longmapsto 5.52\times 10^6$

$\\ \sf\longmapsto 5520000ton$

1ton=1000kg

$\\ \sf\longmapsto 5520000\times 1000$

$\\ \sf\longmapsto 5520000000kg$

$\\ \sf\longmapsto 5.52\times 10^9Kg$

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3 years ago

Read 2 more answers

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

3 years ago