Answer:
I think the second option is correct
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Answer:
<h2>
The magnitude of force F is 18N</h2>
Explanation:
The magnitude of the force in the set up can be solved for using the principle of moment. According to the principle, the sum of clockwise moment is equal to the sum of anticlockwise moments.
Moment = Force * perpendicular distance
Clockwise moments;
The force that acts clockwise is the unknown Force F and 4N force. If the beam rests on a pivot 60 cm from end X and a Force F acts on the beam 80 cm from end X, the perpendicular distance of the force F from the pivot is 80-60 = 20cm and the perpendicular distance of the 4N force from the pivot is 60-50 = 10cm
Moment of force F about the pivot = F * 20
Moment of 4N force about the pivot = 4*10 = 40Nm
Sum of clockwise moment = 40+20F...(1)
Anticlockwise moment;
The 8N will act anticlockwisely about the pivot.
The distance between the 8N force and the pivot is 60-10 = 50cm
Moment of the 8N force = 8*50
=400Nm...(1)
Equating 1 and 2 we have;
40+20F = 400
20F = 400-40
20F = 360
F = 18N
The magnitude of force F is 18N
Answer:
The final equilibrium T_{f} = 25.7[°C]
Explanation:
In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.
That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.
The temperature of thermal equilibrium will be T_f.
The heat absorbed by water will be equal to the heat rejected by Iron.

Heat transfer can be found by means of the following equation.

where:
Qiron = Iron heat transfer [kJ]
m = iron mass = 200 [g] = 0.2 [kg]
T_i = Initial temperature of the iron = 300 [°C]
T_f = final temperature [°C]

Cp_iron = 437 [J/kg*°C]
Cp_water = 4200 [J/kg*°C]
![0.2*437*(300-T_{f})=1*4200*(T_{f}-20)\\26220-87.4*T_{f}=4200*T_{f}-84000\\26220+84000=4200*T_{f}+87.4*T_{f}\\110220 = 4287.4*T_{f}\\T_{f}=25.7[C]](https://tex.z-dn.net/?f=0.2%2A437%2A%28300-T_%7Bf%7D%29%3D1%2A4200%2A%28T_%7Bf%7D-20%29%5C%5C26220-87.4%2AT_%7Bf%7D%3D4200%2AT_%7Bf%7D-84000%5C%5C26220%2B84000%3D4200%2AT_%7Bf%7D%2B87.4%2AT_%7Bf%7D%5C%5C110220%20%3D%204287.4%2AT_%7Bf%7D%5C%5CT_%7Bf%7D%3D25.7%5BC%5D)
Answer:
pressure of the water = 3.3 ×
pa
Explanation:
given data
velocity v1 = 1.5 m/s
pressure P = 400,000 Pa
inside radius r1 = 1.00 cm
pipe radius r2 = 0.5 cm
h1 = 0 (datum at inlet)
h2 = 5.0 m (datum at inlet)
density of water ρ = 1000 kg/m³
to find out
pressure of the water
solution
we consider here flow speed in bathroom that is = v2 and Pressure in bathroom is = P2
here we will use both continuity and Bernoulli equations
because here we have more than one unknown so that
v1 × A1 = v2 × A2 × P1 + ρ g h1 + (0.5)ρ v1² = P2 + ρ g h2 + (0.5) ρ v2²
now we use here first continuity equation for get v2
v2 =
v2 =
v2 = 6 m/s
and now we use here bernoulli eqution for find here p2 that is
P2 = P1 - 0.5× ρ ×(v2² - v1²) - ρ g (h2- h1 )
P2 = 400000 - 0.5× 1000 ×(6² - 1.5²) - 1000 × 9.81 × (5-0 )
P2 = 3.3 ×
pa
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