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Keith_Richards [23]
3 years ago
11

Please help me find the answers!

Physics
1 answer:
VashaNatasha [74]3 years ago
7 0

Answer:

1. T₁ is approximately 100.33 N

T₂ is approximately -51.674 N

2. 230°F is 383.15 K

3. Part A

The total torque on the bolt is -4.2 N·m

Part B

Negative anticlockwise

Explanation:

1. The given horizontal force = 86 N

The direction of the given 86 N force = To the left (negative) and along the x-axis

(The magnitude and direction of the 86 N force = -86·i)

The state of the system of forces = In equilibrium

The angle of elevation of the direction of the force T₁ = 31° above the x-axis

The direction of the force T₂ = Downwards, along the y-axis (Perpendicular to the x-axis)

Given that the system is in equilibrium, we have;

At equilibrium, the sum of the horizontal forces = 0

Therefore;

T₁ × cos(31°) - 86 = 0

T₁ = 86/(cos(31°)) ≈ 100.33

T₁ ≈ 100.33 N

Similarly, at equilibrium, the sum of the vertical forces = 0

∴ T₁×sin(31°) + T₂ = 0

Which gives;

100.33 × sin(31°) + T₂ = 0

T₂ = -100.33 × sin(31°) ≈ -51.674

T₂ ≈-51.674 N

2. 230° F to Kelvin

To convert degrees Fahrenheit (°F) to K, we use;

Degrees \ in  \ Kelvin, K = (x^{\circ} F + 459.67) \times \dfrac{5}{9}

Pluggining in the given temperature value gives;

Degrees \ in  \ Kelvin, K = (230^{\circ} F + 459.67) \times \dfrac{5}{9} = 383.15

230°F = 383.15 K

3. Part A

Torque = Force × perpendicular distance from the line of action of the force

Therefore, the clockwise torque = 9 N × 0.4 m = 3.6 N·m (clocwise)

The anticlockeisre torque = 13 N × 0.6 m = 7.8 N·m (anticlockwise)

The total torque o the bolt = 3.6 N·m - 7.8 N·m = -4.2 N·m (clockwise) = 4.2 N·m anticlockwise

Part B

The torque is negative anticlockwise.

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Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

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     So

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Now the downward normal force acting on the bob is  mathematically represented as

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The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

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     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

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The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

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3 years ago
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