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2 years ago

7

A fault in which the hanging wall moves down relative to the footwall is a _____.

2 answers:

Hoochie [10]2 years ago

8 0

<u>Normal Fault</u> is your answer :)

Have a great day!!!

LuckyWell [14K]2 years ago

4 0

<span>Normal fault. Normal (extensional ) fault is a displacement of a rock as a result of rock-mass movement and occurs when the crust is stretching. Because of the stretching the thickness of the crust is reduced and the crust or horizontally extended. </span>

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Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

sergiy2304 [10]

Answer:

10.53m/s²

Explanation:

Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:

$a = \frac{v^2}{r}$

v is the velocity of the car = 24.5m/s

r is the radius of the track = 57.0m

Substitute the given values into the formula:

$a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}$

Hence the centripetal acceleration of the race car is 10.53m/s²

6 0

3 years ago

An engine has a hot-reservoir temperature of 980 K and a cold-reservoir temperature of 570 K. The engine operates at three-fifth

hram777 [196]

Answer:

efficiency of engine is 0.21

Explanation:

given:

Th=980 K

Tc=570 K

since efficiency of engine is given as:

n=3/5*n_m

=3/5(1-Tc/Th)

=0.21

efficiency of engine is 0.21

7 0

3 years ago

Read 2 more answers

Point charges of 28.0 µC and 42.0 µC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

olga_2 [115]

Answer: a) electric field will be zero at zero meters apart

b) for smaller charge q, E = 6.048X10^6N/m towards away from the charge,

for bigger charge Q, E = 4.032X10^6N/m

Explanation:

Detailed explanation and calculation is shown in the image below.

6 0

3 years ago

The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of

Aleks04 [339]

Answer:

(a) 1, average velocity = -65.6 m/s

2, average velocity = -64.8 m/s

3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given by;

$y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}$

(1)

$y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s$

(2)

$y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s$

(3)

$y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s$

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

5 0

3 years ago

The red lines on this

tino4ka555 [31]

The circular lines you see on the chart are isobars, which join areas of the same barometric pressure.

8 0

2 years ago