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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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If 31.25
valentina_108 [34]
<span>1 C = 6.24150965(16)×10^18 electrons

31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs

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4 0
3 years ago
Read 2 more answers
n a level football field a football is projected from ground level. It has speed 9.0 m/sm/s when it is at its maximum height. It
Pani-rosa [81]

Answer:

T = 0.225 s

Explanation:

The speed of a projectile at the highest point of its motion is the horizontal speed of the projectile. Considering the horizontal motion with negligible air resistance, we can use the following formula:

v_x = RT\\\\T = \frac{v_x}{R}

where,

T = Total time of ball in air = ?

R = Horizontal distance covered = 40 m

v_x = horizontal speed = 9 m/s

Therefore,

T = \frac{9\ m/s}{40\ m}

<u>T = 0.225 s</u>

4 0
3 years ago
Which type of muscle cell can have multiple nuclei
s344n2d4d5 [400]

Answer:

Skeletal muscle cells

Explanation:

Skeletal muscle cells are long, cylindrical, and striated. They are multi-nucleated meaning that they have more than one nucleus. This is because they are formed from the fusion of embryonic myoblasts.

pls mark me the brainliest

7 0
2 years ago
Read 2 more answers
A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk
Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration\frac{m}{s^{2} }

at: truck acceleration\frac{m}{s^{2} })

ac = at= \frac{vf-vi}{t-ti}  equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 \frac{m}{s}, t = final time =5 s

We replaced the known information in the equation(1):

ac = at = \frac{13-0}{15-0}

ac=ac=\frac{13}{15}  \frac{m}{s}

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

Wc=mc*g

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 \frac{m}{s^{2} }

Wc= 2230*9.8

Wc=21854 N

Normal force calculation:

Newton's first law

sum Fy= 0

N-W=0

N=W

N=21854 N

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Ff=0.373*21854

Ff=8151.54 N

Calculation of the tension force in the rope (T):

Newton's Second law

sum Fx= mc*ac

T-Ff=mc*ac

T=2230(\frac{13}{15}) + 8151.54

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0
3 years ago
Convert 123,453 to a scientific notation
Natalija [7]

Answer:

1.23453*10^5 is scientific way

3 0
2 years ago
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