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nirvana33 [79]
3 years ago
9

A 100.0-kg bakery sign hangs from two thin cables as shown.

Physics
1 answer:
harina [27]3 years ago
5 0

Answer:

1. T₁ = 500 N

2. T₂ = 866 N

Explanation:

Please see attached photo for the diagram.

Thus, we can obtain obtained the value of T₁ and T₂ as follow:

1. Determination of T₁

Angle θ = 30

Hypothenus = 100 kg

Opposite = T₁ =?

Sine θ = Opposite /Hypothenus

Sine 30 = T₁ / 100

Cross multiply

T₁ = 100 × Sine 30

T₁ = 100 × 0.5

T₁ = 50 Kg

Multiply by 10 to express in Newton

T₁ = 50 × 10

T₁ = 500 N

2. Determination of T₂

Angle θ = 60

Hypothenus = 100 kg

Opposite = T₂ = ?

Sine θ = Opposite /Hypothenus

Sine 60 = T₂ / 100

Cross multiply

T₂ = 100 × Sine 60

T₂ = 100 × 0.8660

T₂ = 86.6 Kg

Multiply by 10 to express in Newton

T₂ = 86.6 × 10

T₂ = 866 N

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dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

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    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

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5 0
3 years ago
How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v
777dan777 [17]
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
W= qV_i - qV_f
where q is the charge of the proton, q=1 e = 1.6\cdot 10^{-19}C, with e being the elementary charge, and V_i = +125 V and V_f = -55 V are the initial and final voltage.

Substituting, we get (in electronvolts):
W=e(125 V-(-55 V))=180 eV
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3 years ago
A meter stick balances horizontally on a knife-edge at the 51 cm mark. With two nickels stacked over the 6.0 cm mark, the stick
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Answer:

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Explanation:

Two main conditions for equilibrium are:

I. The resultant force must be equal to zero. That is, sum of the forces acting in one direction about a point must be equal to the sum of the forces acting in the opposite direction about the same point.

II. The resultant moment must be equal to zero. That is, sum of the moments in one direction about a point must be equal to the sum of the moments in another direction about the same point.

For the above question,

the 51cm mark is the point where the resultant weight of the meter stick lies,

the pivot or point is the 45cm mark where the stick balanced when 2 nickels ( total mass (5.0g x 2) 10g were placed at the 6cm mark.

Using the conversion factor:

1000g(1kg) = 10N, we can convert mass to weight, calculate the weight of the meter stick then reconvert to mass.

That is,

mass of 2 nickels = 10g = 10/1000 = 0.01N.

Moment = Force x distance from line of force to pivot of rotation

Applying the principle of equilibrium,

Moment of left side = Moment of right side

0.01 x (45-6) = W x (51-45)

Where W = weight of the meter stick

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W x 6 = 0.39

W = 0.39/6

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zhenek [66]

Answer:

9.4

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