Answer:
b)1.08 N
Explanation:
Given that
velocity of air V= 45 m/s
Diameter of pipe = 2 cm
Force exerted by fluid F

So force exerted in x-direction


F=0.763 N
So force exerted in y-direction


F=0.763 N
So the resultant force R


R=1.079
So the force required to hold the pipe is 1.08 N.
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
Answer:
The white lined paper
Explanation:
The teacher is most likely putting the while line paper in jeopardy because of the detail process involved in taking care of the paper prior to the submission of the home work.
The fact that a mistake must not be visible due to the instruction of every erasures being thorough and clean. this can cause jeopardy to the paper.
Structure Of The Atom: Our current model of the atom can be broken down into three constituents parts – protons, neutron, and electrons. Each of these parts has an associated charge, with protons carrying a positive charge, electrons having a negative charge, and neutrons possessing no net charge.
Answer:
<em>The temperature will be greater than 25°C</em>
Explanation:
In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.
mathematically
Change in the internal energy of a system ΔU = ΔQ + ΔW
in an adiabatic process, ΔQ = 0
therefore
ΔU = ΔW
where ΔQ is the change in heat into the system
ΔW is the work done by or done on the system
when work is done on the system, it is conventionally negative, and vice versa.
also W = pΔv
where p is the pressure, and
Δv = change in volume of the system.
In this case,<em> work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C </em>