Answer: 14.52*10^6 m/s
Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.
the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:
e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s
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:
Water flowing rate= (300000kg/s) = (300000l/s)
Explanation:
First with the section of the channel, the depth of the water and the speed of the fluid we can determine the volume of fluid that circulates per second through the channel:
Volume per time= 15m × 8m × (2.5m/s)= 300 m³/s
With this volume of circulating fluid per second elapsed, we multiply it by the density of the water to determine the kilograms or liters of water that circulate through the channel per second elapsed:
Water flowing rate= (300m³/s) × (1000kg/m³)= (300000kg/s) = (300000l/s)
Taking into account that 1kg of water is approximately equal to 1 liter of water.
