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GenaCL600 [577]
2 years ago
6

1. Mike wears an 8.50 kg backpack while scaling a cliff. After 90 minutes, Mike is 11.2 m above the starting point.

Physics
1 answer:
svet-max [94.6K]2 years ago
8 0

Answer:

a. 952Joules

b. 8008 joules

c. 1.48 watts

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Compared to the sun, a star whose spectrum peaks in the infrared is:
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It is impossible to destroy __________.
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Read 2 more answers
A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo
m_a_m_a [10]

Answer:

The induced emf  is  \epsilon  = 0.1041 \  V  

Explanation:

From the question we are told that

   The  radius of the circular loop is  r =  9.50 \ cm  =  0.095 \ m

     The  intensity of the wave is  I  =  0.0215 \ W/m^2

      The wavelength is  \lambda =  6.90\ m

Generally the intensity is mathematically represented as

         I  =  \frac{ c *  B^2  }{ 2 * \mu_o  }

Here  \mu_o is the permeability of free space with value  

         \mu_o  =  4 \pi *10^{-7} N/A^2

B is the magnetic field which can be mathematically represented from the equation as

          B  =  \sqrt{ \frac{ 2 *  \mu_o  *  I  }{ c} }

substituting values

          B  =  \sqrt{ \frac{ 2 *  4\pi *10^{-7} *   0.0215  }{ 3.0*10^{8}} }

          B  =  1.342 *10^{-8} \  T

The  area is mathematically represented as

       A =  \pi r^2

substituting values

       A =  3.142 *   (0.095)^2

       A = 0.0284

The angular velocity is mathematically represented as

        w =  2 *  \pi  *  \frac{c}{\lambda }

substituting values          

       w =  2 *  3.142   *  \frac{3.0*10^{8}}{ 6.90 }  

        w =  2.732 *10^{8} rad  \ s^{-1}  

Generally the induced emf is mathematically represented as

        \epsilon  =  N *  B  *  A  *  w * sin (wt )

At maximum induced emf  sin (wt)  =  1

    So

         \epsilon  =  N *  B  *  A  *  w

substituting values

         \epsilon  = 1  *    1.342 *10^{-8}   *  0.0284  *2.732 *10^{8}  

         \epsilon  = 0.1041 \  V  

         

7 0
3 years ago
Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm
DENIUS [597]

data which is expressed in form of following way

a = a_o + \Delta a

here in above expression

a_o = true value

\Delta a = uncertainty in the value

now the relative uncertainty is given as

\frac{\Delta a}{a_o}

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = \frac{0.05}{2.70} = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = \frac{0.08}{12.02} = 0.00665

4 0
3 years ago
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