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svp [43]
2 years ago
8

What is the net force (including direction and magnitude) acting on the object on the above left?

Physics
1 answer:
Ber [7]2 years ago
7 0

Answer:

Explanation:

the net force on the right left is 25 N and is directed upward

the net force on the left one is zero because 200N force act upward and 200N force act downward so both cancel each other and net force is zero

i hope this will help you

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An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.
jekas [21]

Answer:

\varphi_1= 796.25 N m^2/C

\varphi_2= 0 N m^2/C

\varphi_3=686.1  N m^2/C

Explanation:

From the question we are told that

Electric field of intensity E= 3.25 kN/C

Rectangle parameter Width W=0.350 m  Length L=0.700 m

Angle to the normal \angle=30.5 \textdegree

Generally the equation for Electric flux at parallel to the yz plane \varphi_1 is mathematically given by

\varphi_1=EA cos theta

\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0

\varphi_1= 796.25 N m^2/C

Generally the equation for Electric flux at parallel to xy  plane \varphi_2 is mathematically given by

\varphi_2=EA cos theta

\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90

\varphi_2= 0 N m^2/C

Generally the equation for Electric flux at angle 30 to x plane \varphi_3 is mathematically given by

\varphi_3=EA cos theta

\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5

\varphi_3=686.072219  N m^2/C

\varphi_3=686.1  N m^2/C

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3 years ago
Dolphins communicate using compression waves (longitudinal waves).  Some of the sounds dolphins make are outside the range of hu
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The dolphin would send an underwater message because the water is denser than air and therefore the sound would be louder and would travel faster. The denser the object the faster the sound travels.
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3 years ago
Read 2 more answers
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

     I  =  J*A

=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

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