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zheka24 [161]
3 years ago
15

Part A What is the resistance of a 4.4 m length of copper wire 1.3 mm n diameter? The resistivity of copper is 1.68x 10-8 Ω-m Ex

press your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Natalija [7]3 years ago
7 0

Explanation:

It is given that,

Length of the copper wire, l = 4.4 m

Diameter of copper wire, d = 1.3 mm = 0.0013 m

Radius of copper wire, r = 0.00065 m

The resistivity of the copper wire, \rho=1.68\times 10^{-8}\ \Omega-m

We need to find the resistance of the copper wire. It is given by :

R=\rho\dfrac{l}{A}

R=1.68\times 10^{-8}\ \times \dfrac{4.4\ m}{\pi (0.00065)^2}

R =0.055 ohms

So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.

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588 N

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Conveyor belts are often used to move packages around warehouses. The conveyor shown below moves packages at a steady 4.0 m/s. A
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A ____________ sloped line means the object is returning to the starting point. A Upward B Backwar C Missing D Downward
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Answer:

A downward sloped line means the object is returning to the starting point.

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3 years ago
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

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

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

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

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

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

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     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

    T_x  = 0.3294 \ N

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

         

3 0
3 years ago
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lara [203]

Answer:

The cooling time will not be reduced.

Explanation:

The time to cook is virtually the same in both types, vigorously and gently boiling water.

The reason cooking of spaghetti  calls for vigorously boiling water is to keep the pasta agitated so that they do not stick to one another.

The temperature of boiling water is the same for both vigorously boiling water and gently boiling water, therefore there will be little time difference in when the potatoes will cook when it is done with vigorously boiling water than when it is cooked with gently boiling water.

However cooking potatoes in vigorously boiling water may cause the water to dry up on time and the potatoes get burnt.

8 0
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