Here is the correct answer of the given problem above.
Given that the basket has a mass of 5.5kg, the magnitude of the normal force if the basket is at rest on a ramp inclined above the horizontal is at 12 degrees. The solution is simple:
<span>Fn at rest = lmgl </span>
<span>= 5.5kg (9.80N/kg)
=</span><span> mgCos12degrees
Hope this answer helps. </span>
Power = Work / time
The work given here is 83J and the time it took to do 83J of work was 3s
So..
Power = 83J / 3s
Power = 27.67 W or 27.7 W
The valence electrons are the one furthest from the nucleus
The correct answer is D.
A nucleon<span> is one of either of the two types of subatomic particles (neutrons and protons) which are located in the nucleus of atoms.
</span>
The total number of nucleon in the nucleus of an atom gives you an idea about the mass of that atom. In fact, one may refer mass number as nucleon number.
Simply put, nucleons are the particles that make nucleus of an atom and are held up together inside the nucleus due to nuclear force.
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
