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zlopas [31]
2 years ago
14

How do I calculate equilibrant and fx and fy. I don't understand what they are asking

Physics
1 answer:
Natasha2012 [34]2 years ago
3 0

(a) The equilibrant C for force of vector A and B is 3.43 N.

(b) The equilibrant C for fx of vector A and B is 2.1 N.

(c) The equilibrant  C, for fy of vector A and B is 2.12 N.

<h3>What is equilibrant force?</h3>

An equilibrant force is a single force that will bring other bodies into equilibrium.

<h3>From configuration 1:</h3>

Vector A: mass = 0.2 kg, θ = 20⁰

Vector B: mass = 0.15 kg, θ = 80⁰

Fx = mg cosθ

Fy = mg sinθ

where;

  • m is mass
  • g is acceleration due to gravity

<h3>Vector A</h3>

Force of A due to its weight

F(A) = mg

F(A) = 0.2 x 9.8 = 1.96 N

Fx = (0.2 x 9.8) cos(20) = 1.84 N

Fy = (0.2 x 9.8) sin(20) = 0.67 N

<h3>Resultant force</h3>

R = √(0.67² + 1.84²)

R = 1.96 N

<h3>Vector B</h3>

Force of B due to its weight

F(B) = mg

F(B) = 0.15 x 9.8

F(B) = 1.47 N

Fx = (0.15 x 9.8) cos(80) = 0.26 N

Fy = (0.15 x 9.8) sin(80) = 1.45 N

<h3>Resultant force </h3>

R = √(0.26² + 1.45²)

R= 1.47 N

<h3>Equilibrant  C of vector A and B</h3>

Equilibrant force:

Force, C = 1.96 N + 1.47 N

Force, C = 3.43 N

Equilibrant FX:

Fx, C = Fx(A) + Fx(B)

Fx, C = 1.84 N + 0.26 N = 2.1 N

Equilibrant FY:

Fy, C = Fy(A) + Fy(B)

Fy, C =0.67 N + 1.45 N = 2.12 N

Learn more about equilibrant force here: brainly.com/question/8045102

#SPJ1

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   w = 0.55 rad / s

Explanation:

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    L₀ = r p + 0

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      Lf = (I + m r²) w

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A beach ball with a volume of 5,000 cm3 is pushed underwater. What is the magnitude of the buoyant force pushing upwards? The de
Kobotan [32]
<h2>Hello!</h2>

The answer is:

The buoyant force is equal to 49N.

<h2>Why?</h2>

The buoyant force is the force that pushes upwards and object when it's submerged in water, this force is always trying to move the object to the surface of the liquid or water.  We must consider that the volume of water or liquid displaced is equal to the volume of the submerged object.

We can calculate the buoyant force using the following formula:

F_{B}=Density*VolumeDisplaced*GravityAcceleration

Where,

Density is the density of the water or liquid.

Volume displaced is equal to the volume of the submerged object.

Gravity acceleration is the acceleration due to gravity.

So, from the statement we know that:

VolumeDisplaced=BeachBallVolume=5000cm^{3}\\\\Density=1\frac{gram}{cm^{3} }=\frac{0.001kg}{cm^{3} }\\\\g=9.8\frac{m}{s^{2} }

Now, substituting and calculating we have:

F_{B}=Density*VolumeDisplaced*GravityAcceleration

F_{B}=0.001\frac{kg}{cm^{3}*} *5000cm^{3}*9.8\frac{m}{s^{2} }

F_{B}=0.001\frac{kg}{cm^{3}*} *5000cm^{3}*9.8\frac{m}{s^{2} }

F_{B}=49\frac{Kg.m}{s^{2}}=49N

Hence, we have that the correct answer is:

The buoyant force is equal to 49N.

Have a nice day!

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