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3 years ago

A ____fault forms as a result of horizontal compression.

Physics

1 answer:

kirill [66]3 years ago

4 0

Answer:

Fracture or system of fractures in Earth's crust that occurs when stress is applied too quickly or stress is too great; can form as a result of horizontal compression (reverse fault), horizontal shear (strike-slip fault), or horizontal tension (normal fault).

Explanation:

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Convert the following.

Alexandra [31]

Answer:

50°C whn converted into K - 323 K

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3 years ago

A 10.0 m wire is hung from a high ceiling and held tightly below

yan [13]

Answer:

5 m

Explanation:

Speed of waves is as the product of frequency and wavelength hence expressed as s=fw where f is the frequency of waves in Hz, s is the speed in m/s and w is wavelength in meters.

Making w the subject of the formula then

$w=\frac {s}{f}$

Substituting 335 m/s for s and 67 Hz for f then the wavelength is

$w=\frac {335}{67}=5m$

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3 years ago

A value that describes how heavy an object is and is related to the force of gravity is:

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The weight of an object

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3 years ago

Read 2 more answers

A weightlifter curls a 30 kg bar, raising it each time a distance of 0.50 m .How many times must he repeat this exercise to burn

ludmilkaskok [199]

Answer:

$N = 2141 times$

Explanation:

Each time the work done to raise a given mass is

$W = mgh$

here we know that

$m = 30 kg$

$h = 0.50 m$

now we have

$W = (30 kg)(9.81 m/s^2)(0.50)$

$W = 147.15 J$

since it is just 25% of actual energy consumed as we know its efficiency is 25%

so we have total energy consumed in this way

$E_{total} = \frac{147.15}{0.25}$

$E_{total} = 588.6 J$

now if it took N number of times so burn the fat of a pizza then

$N(588.6) = 1260 \times 10^3$

$N = 2141 times$

4 0

3 years ago

suppose that the man pictured on the front side is orbiting the earth (mass=5.98 x 10^24kg at a distance of 310 miles above the

wlad13 [49]

Answer:

a = 9.81[m/s^2]; v = 18683.5[m/s]

Explanation:

The stament of the problem is:

Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).

a. What acceleration does he experience due to the earth's pull?

b. What tangential velocity must he possess in order that he orbit safely (in m/s)?

First we need to convert all the initial data to units of the SI

Rs = 310 [mil] = 498897 [m]

RT= 400 [mil] = 643738 [m]

r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"

$F=G*\frac{M*m}{r^{2} } \\where:\\$

G = universal gravitational constant

M = mass of the earth [kg]

m = mass of the man [kg]

r = distance from the center of the earth to the man [m]

The acceleration he is experimenting is the same acceleration given by the gravity, therefore:

a = g = 9.81[m/s^2]

To find the tangential velocity, we must determinate the force exerted by the earth.

Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.

$G = 6.67*10^{-11} [\frac{N*m^{2} }{kg^{2} } ]\\M=5.98*10^{24}[kg]\\ m=100 [kg]\\Replacing:\\F = G*\frac{M*m}{r^{2} }$

$F = 6.67*10^{-11}*\frac{5.98*10^{24}*100 }{1142635^{2} } \\ F= 30550 [N]$

And this force will be equal to the following expression:

$F = m*\frac{v^{2} }{r} \\where:\\v= tangential velocity [m/s]\\v=\sqrt{\frac{F*r}{m} } \\v=\sqrt{\frac{30550*1142635}{100} } \\v=18683.5[m/s]$

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3 years ago