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zzz [600]
2 years ago
14

3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of

37° with the horizontal, as shown. If the cofficient of kinetic friction between the block and the wall is 0.30. Find the work done by the applied force the frictional force.​
Physics
1 answer:
Andrews [41]2 years ago
5 0

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

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3 years ago
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eimsori [14]

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Frequency is the number of waves produced per second. Frequency and wavelength are inversely proportional .Thus, if the frequency is doubled the wavelength would be halved.


6 0
3 years ago
3.00 m^3 of water is at 20.0°C.
krok68 [10]

Answer:

9m^3

Explanation:

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volume  v2=  ???

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V1/T1= V2/T2

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6 0
3 years ago
n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet
enot [183]

Answer:

The  corresponding  magnetic field is  

Explanation:

From the question we are told that

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Generally the  magnetic  field amplitude is  mathematically represented as

              B_o  =  \frac{E_o }{c }

Where c is the speed of light with a constant value

         c = 3.0 *0^{8} \ m/s

So  

        B_o   =  \frac{611 }{3.0*10^{8}}

         B_o   =  2.0 4 *10^{-6} \  Vm^{-2} s

Since 1  T  is  equivalent to  V  m^{-2} \cdot  s

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6 0
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Anna007 [38]

Answer:

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The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.

Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:

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This corresponds to the second option shown in the question: "Voltage times amperage".

6 0
4 years ago
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