A thrust fault is a reverse fault with an extremely high dip (close to 90°). This is the false statement.
Answer: Option D
<u>Explanation:</u>
Faults are the fracture or fracture zone occurring on the rocks. These fractures can travel through the rocks leading to massive destruction. So, depending upon the direction of their travel, the faults can be classified as normal, reverse and strike slip fault. Also, the angle of dip along the fault is one of the important criteria for determining the type of faults.
There is dip-slip fault which has its movement along the vertical fault plane while the strike slip fault will be in horizontal direction. Similarly, an oblique fault will be acting in both vertical and the horizontal direction. So, the fourth statement related to thrust fault is false as in reverse fault or thrust fault the dip will be shallow and not high.
Answer:
option B.
Explanation:
The correct answer is option B.
The phenomenon of the curtains to pull out of the window can be explained using Bernoulli's equation.
According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.
When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.
Answer:
<h2>5.6kW</h2>
Explanation:
Step one:
given
mass m= 24kg
distance moved= 6m
time taken= 4seconds
Step two:
Required
power
but work done is the force applied at a distance, and the power is the work done time the time taken
Work done= F*D
F=mg
W= mg*D
W=24*9.81*6
W=1412.6J
Power P= work * time
P=1412.6*4
p=5650.5W
P=5.6kW
Refer to the diagram shown below.
Neglect wind resistance, and use g = 9.8 m/s².
The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²
Answer: - 82 m/s² (or a deceleration of 82 m/s²)
Answer:
The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

Explanation:
From the question we are told that
The mass of the rocket is m = 6300 kg
The velocity at gas is being ejected is u = 2000 m/s
The initial acceleration desired is 
The time taken for the gas to be ejected is t = 1 s
Generally this desired acceleration is mathematically represented as

Here
is the rate at which gas is being ejected with respect to time
Substituting values

=> 
=> 
=> 
=> 