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3 years ago

A fault that is formed when compression causes the hanging wall to move over the foot wall is a(n) ______.

2 answers:

4 0

A fault that is formed when compression causes the hanging wall to move over to the foot wall is a : Reverse Fault
Reverse fault is exactly the opposite of the normal fault which will occur on the region/area undergoing the compression

hope this helps<span />

Fynjy0 [20]3 years ago

3 0

It is Reverse or Reverse fault in other words - I just took the test this was the answer.

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What type of triangle measures 8 inches,8 inches,3 inches

kykrilka [37]

Answer:

isoceles

Explanation:

Because two sides are equal

7 0

3 years ago

Read 2 more answers

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

The microwaves in a certain microwave oven have a wavelength of 12.2 cm. How wide must this oven be so that it will contain five

leva [86]

Answer:

$l = 0.305 \ m$

$f = 3.0*10^{11} \ Hz$

Explanation:

From the question we are told that

The wavelength is $\lambda = 12.2 \ cm = 0.122 \ m$

The number of antinodal planes of the electric field considered is n = 5

The width is mathematically represented as

$l = \frac{ n \lambda}{2}$

$l = \frac{5 * 0.122 }{ 2}$

$l = 0.305 \ m$

Generally the frequency the errors was made is mathematically represented as

$f = \frac{c}{\lamda_k}$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

$\lambda_k$ is the wavelength of the microwave has to be in order for there still to be five antinodal planes of the electric field along the width of the oven, which is mathematically represented as

$\lambda_k = \frac{ \lambda * \frac{0.04}{2} }{n/2}$

$\lambda_k = \frac{0.122*0.02}{5/2}$

$f = \frac{3.0*10^{8}}{0.000976}$

$f = 3.0*10^{11} \ Hz$

8 0

3 years ago

Urgent!!!!!! 20 points!!!!!!<br> how are the speed, wavelength and frequency of a wave related

zheka24 [161]

Answer:

The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. Speed shows how long it takes for wavelengths to travel.

6 0

3 years ago

A solid nonconducting sphere of radiusRcarries a chargeQdistributed uniformly throughout itsvolume. At a certain distancer1(r1&l

lara [203]

Answer:

$E' = \frac{1}{8} E$

Explanation:

Given data:

first case

Distance of electric field from center of sphere is r_1 <R

Electric field at r_1< R

$E = \frac{kQr_1}{R^3}$

second case

Distance of electric field from centre of sphere is r_1 < 2R

Electric field at r_1< 2R

$E' = \frac{kQr_1}{8R^3}$

so, we have

$E' = \frac{1}{8} E$

3 0

2 years ago