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saul85 [17]
3 years ago
6

A uniform shelf 1.8 meters long with mass 15 kg is mounted by a small hinge on a wall. The beam is held in a horizontal position

by a cable that makes an angle of 37°. There is a 4.9-kg round light hanging down and is 0.55 meters from the wall. What is the tension in the supporting cable?
Physics
1 answer:
jeka57 [31]3 years ago
8 0

Answer:

 T = 153.72 N

Explanation:

For this exercise we must use the conditions of translational and rotational equilibrium.

Let's set a frame of reference on the hinge, start by writing the rotational equilibrium relationship, suppose counterclockwise rotation is positive

We look for the components of the cable tension with trigonometry

           cos 37 = Tₓ / T

           sin 37 = T_{y} / T

           Tₓ = T cos 37

           T_{y} = T sin 37

the expression for rotational equilibrium is

        T_{y} L + Tₓ 0 - W L / 2 - W_light  0.55 = 0

where L is the length L=  1.8 m,

        T_{y} = (W L/2 + W_lght 0.55) / L

        T sin 37 = Mg /2 + m_light g 0.55 / L

        T = (M / 2 + m_light 0.55 / L) g / sin 35

 let's calculate

        T = (15/2 + 4.9 0.55 / 1.8) 9.8 / sin 35

         T = 153.72 N

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Fossil fuels have led to an increase in food costs.

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3 years ago
______ 4. Justin was creating a chart comparing the potential energy between charged particles and the potential energy of magne
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Answer:

The correct option is;

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Here we have that when we move the like poles of two bar magnets close to each other, there is an increased resistance in the continuing motion, therefore for each extra gap closer achieved, there is an increase in potential energy

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A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C
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Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

d_{ab} = distance traveled from station A to station B

t_{ab} = time of travel between station A to station B

we know that

Time = \frac{distance}{speed}

t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}

Total distance traveled is given as

d = d_{ab} + d_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }

Given that :

d_{ab} = 4 d_{bc}

So

v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}

2)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

t_{ab} = time of travel between station A to station B

d_{ab} = distance traveled from station A to station B

we know that

distance = (speed) (time)

d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = time of travel for train from station B to station C

we know that

distance = (speed) (time)

d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}

Total distance traveled is given as

d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}

Given that :

t_{ab} = 3 t_{bc}

So

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