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Vanyuwa [196]
3 years ago
15

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon

has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.4 m/s . What upward force must a branch provide to support the swinging gibbon?
Physics
1 answer:
Katarina [22]3 years ago
7 0

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

F_{branch} + F_{gravity}=F_{centripetal}

force of gravity = - mg

F_{branch}=F_{centripetal}-F_{gravity}

                        = \dfrac{mv^2}{r} + mg

                        = m(\dfrac{3.4^2}{0.6} +9.8)

                        =9 x 29.067

                        = 261.6 N

upward force acting = 261.6 N

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The resistance of a single light bulb is 220 ohms per bulb.

<h3>What is Ohm's Law?</h3>

Ohm's Law is a formula used to determine how voltage, current, and resistance in an electrical circuit relate to one another.

Ohm's Law (E = IR) is as basic to students of electronics as Einstein's Relativity equation (E = mc2) is to physicists.

E = I x R

The formula reads voltage = current x resistance, or V = A xΩ., or volts = amps x ohms.

110volts divided by .25amps = 440 ohms. 440 divided by 2 =220 ohms per bulb.

R = 110/(2*0.25) = 220 ohms

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2 years ago
Why is an iron bolt attracted to a magnet?
icang [17]

Answer:

A. When it is in a magnetic field, it becomes a temporary magnet.

Explanation:

An iron bolt is attracted to a magnet because when in a magnetic field, the iron becomes a temporary magnet.

This is because the iron aligns their electrons in the magnetic fields.

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8 0
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A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force act
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Answer:

cos 0 = 1.

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Explanation:

7 0
3 years ago
Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?
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The equilibrium conditions allow to find the results for the balance forces are:

  • F₁ = 225.4 N
  • F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

            ∑ F = 0

            ∑ τ = 0

           

Where F are the forces and τ the torques.

The torque  is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

           F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

         F₂ 2 - W₁ 1 - W₂ 1.5 = 0\frac{W_1  \ 1 + W_2 \ 1.5}{2}

Let's calculate F₂

         F₂ = \frac{W_1 \ 1 + W_2 \ 1.5 }{2}  

         F₂ = (m g + M g 1.5)/ 2

         F₂ = \frac{(12 + 68 \ 1.5 ) \  9.8}{2}  

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We substitute in the translational equilibrium equation.

         F₁ = W₁ + W₂ - F₂

         F₁ = (m + M) g - F₂

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In conclusion using the equilibrium conditions we can find the forces of the balance are:

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a = 1.5 m/sec²

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