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kirill115 [55]
3 years ago
7

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

Physics
2 answers:
Hoochie [10]3 years ago
8 0

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

Have a great day!!!

LuckyWell [14K]3 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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How are theories and laws connected ​
Anni [7]

Answer:

Laws are statements about something that's been observed and stated while a theory is an explanation of what's been observed. This connection between them forms a main idea that many people regulate as "what's normal."

Explanation:

4 0
3 years ago
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What is the momentum of a 1550kg car that is traveling 38.0 m/s?
konstantin123 [22]

Answer:

p = 58,900 kg m/s

Explanation:

p = m × v

p = 1,550 × 38.0

p = 58,900 kg m/s

5 0
2 years ago
During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc
Lina20 [59]
V o = 6 m/s,
t = 2 s
v = 10 m/s
v = v o + a t
a t = v - v o
a = ( v - v o ) / t 
a = ( 10 m/s - 6 m/s ) / 2 s = 4 m/s / 2 s = 2 m/s²
Answer:
The runner`s acceleration is 2 m/s².
6 0
3 years ago
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the
marusya05 [52]

Answer:

Explanation:

Given

Initial velocity of ball u=10\ m/s

height of window h=20\ m

Using Equation of motion

y=ut+\frac{1}{2}at^2

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

Y=ut+\frac{1}{2}at^2+20

Y=10\times t+0.5\times (-9.8)t^2+20

Y=-4.9t^2+10t+20

(b)highest point is obtained at v=0

v^2-u^2=2as

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

(0)-10^2=2\times (-9.8)\times s

s=\frac{100}{19.6}

s=5.102\ m

Highest Point will be s+20=25.102\ m

(c)Time taken when the ball hit the ground i.e. at Y=0

-4.9t^2+10t+20=0

t=3.28\ s

impact velocity v=\sqrt{2\times 9.8\times 25.102}

v=22.181\ m/s

7 0
3 years ago
Two rollerbladers face each other and stand at rest on a flat parking lot. Tracey has a mass of 32 kg, and Jonas has a mass of 4
satela [25.4K]

Mass of Tracey M1 = 32 kg

Mass of Jonas M2 = 45 kg

Initially both were at rest

so V1i = V2i =0

after pushing each other Jonas speed V2f = 0.80 m/s

we need to find out final speed of Tracy

Here we can use momentum conservation as no external force is acting here

M1V1i + M2V2i = M1V1f + M2V2f

32(0) + 45(0) = 32 V1f + 45(0.80)

0 = 32 V1f + 36

-36 = 32 V1f

V1f = - 1.125 m/s

negative sign shows that Tracy will move opposite to the Jonas

so answer in two significant figure would be

V1f = 1.1 m/s

7 0
3 years ago
Read 2 more answers
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