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Final answer:
This Physics problem involves balancing the forces and torques acting on a 3.6-m-long pole. By applying the principles of equilibrium and calculations of torque, we find that 114 N of force is needed to keep the pole in a horizontal position.
Explanation:
This is a physics problem related to the concepts of equilibrium and torque. From the details provided, we know that the pole has a mass of 21 kg and it's 3.6 meters long. The center of gravity (cg) of the pole, since it's uniform, is at the middle, which is at 1.8 m from either end of the pole. We are then told that you are holding the pole 35 centimeters (or 0.35 meters) from its tip.
To keep the pole horizontal in equilibrium, the downward force due to the weight of the pole at its center of mass (which is equal to the mass of the pole times gravity, or 21*9.8 = 205.8 N) needs to be balanced by the sum of the torques produced by the forces you are applying at the end you are holding and the force exerted by the fence post at the other end.
Let the force you apply be F1 and the force the fence post exerts be F2. We have F2 at 0.35 m from one end (the pivot point), and F1 at 3.6 - 0.35 = 3.25 m from the pivot. Given that the torque (t) equals to Force (F) times the distance from the pivot (d), and that the net torque should equal zero in equilibrium, we have:
0.35*F2 = 3.25*F1 (1)
Because the net force should also be zero in equilibrium, we have:
F1 + F2 = 205.8 (2)
Solving these two equations, we'll be able to calculate that the force you must exert to keep the pole motionless in a horizontal position, F1, is approximately 114 N.
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Final answer:
To balance the 3.6m-long, 21 kg pole and keep it horizontally motionless, a force of approximately 114N is required
Explanation:
The subject question is a classic example of Torque problem specific to Physics, which involves the concepts of force, weight, and distance. To keep the pole motionless and horizontally balanced, the force you exert must counterbalance the torque due to the pole's weight. Assuming the pole is uniform, its center of gravity (cg) is at its midpoint, 1.8m from each end. The weight of the pole acts downward at this midpoint, providing a clockwise torque about the point of support, which is the fence post.
This torque is calculated as Torque = r * F = 1.8m (distance from fence post to cg) * Weight of pole = 1.8m * 21kg * 9.8m/s² (gravitational acceleration) = ~370 N.m. As the pole is motionless, the total torque about any point must be zero. Hence, the counter-clockwise torque provided by the force you exert is equal to the clockwise torque due to the weight of the pole. Using the distance from the point of your hold to the fence post (3.25m) we can calculate the force you need to exert: Force = Torque/distance = 370 N.m/3.25m = ~114N.
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Answer:
128 s
Explanation:
The distance, speed and time are related as;
Given that the speed = 5 m/s
Distance = 640 m
Time = ?
So,
Thus, Garza takes 128 s to travel 640 m at 5 m/s speed.
Answers
Answer:
m2=3.2722lbm/s
Explanation:
Hello!
To solve this problem follow the steps below
1. Find water densities and entlapies in all states using thermodynamic tables.
note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)
through prior knowledge of two other properties, such as pressure and temperature.
D1=Density(Water;T=50;x=0)=62.41 lbm/ft^3
D2=Density(Water;T=120;x=0)=61.71 lbm/ft^3
D3=Density(Water;T=80;x=0)=62.21 lbm/ft^3
h1=Enthalpy(Water;T=50;x=0)=18.05 BTU/lbm
h2=Enthalpy(Water;T=120;x=0)=88 BTU/lbm
h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm
2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit
m1+m2=m3
3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out
m1h1+m2h2=m3h3
18.05(m1)+88(m2)=48.03(m3)
divide both sides of the equation by 48.03
0.376(m1)+1.832(m2)=m3
4. Subtract the equations obtained in steps 3 and 4
m1 + m2 = m3
-
0.376m1 + 1.832(m2) =m3
--------------------------------------------
0.624m1-0.832m2=0
solving for m2
(0.624/0.832)m1=m2
0.75m1=m2
5. Mass flow is the product of density by velocity across the cross-sectional area
m1=(D1)(A)(v1)
internal Diameter for 2" Sch 40=2.067in=0.17225ft
m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s
6.use the equation from step 4 to find the mass flow in 2
0.75m1=m2
0.75(4.3629)=m2
m2=3.2722lbm/s
Answers
Answer:
Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.
Explanation:
Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.
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Answer:
What displacement must the physics professor give the car
= 12.91 METERS
Explanation:
Check the attached file for explanation
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Answer:
a. 409.5 m/s b. f₁ = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz
Explanation:
a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³
So, v = √(B/ρ)
= √(1.09 × 10⁷ N/m²/65.0 kg/m³)
= √(0.01677 × 10⁷ Nm/kg)
= √(0.1677 × 10⁶ Nm/kg)
= 0.4095 × 10³ m/s
= 409.5 m/s
b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m
Now, for the first mode or frequency, n = 1
f₁ = v/4L
= 409.5 m/s ÷ (4 × 0.75 m)
= 409,5 m/s ÷ 3 m
= 136.5 Hz
Now, for the second mode or frequency, n = 2
f₂ = 3v/4L
= 3 ×409.5 m/s ÷ (4 × 0.75 m)
= 3 × 409,5 m/s ÷ 3 m
= 3 × 136.5 Hz
= 409.5 Hz
Now, for the third mode or frequency, n = 5
f₃ = 5v/4L
= 5 × 409.5 m/s ÷ (4 × 0.75 m)
= 5 × 409,5 m/s ÷ 3 m
= 682.5 Hz