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timofeeve [1]
3 years ago
11

When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a f

orce of 89 N, what is the magnitude of the horizontal component of the force that the floor exerts on the block?
Physics
1 answer:
elena-s [515]3 years ago
3 0

Answer:

Explanation:

When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .

It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

= 89 - 76

= 13 N

Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )

=

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g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 × 1026 W and assume that th
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Answer:

The value is N  =  1.107 *10^{45 }  \ photons    

Explanation:

From the question we are told that

   The  power output from the sun is  P_o =  4 * 10^{26} \  W

   The average wavelength of each photon is  \lambda  = 550 \  nm  =  550 *10^{-9} \  m

Generally the energy of each photon emitted is mathematically represented as

        E_c =  \frac{h * c  }{ \lambda }

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          c is the speed of light with value  c =  3.0 *10^{8} \  m/s

So

       E_c =  \frac{6.62607015 * 10^{-34}  * 3.0 *10^{8}  }{ 550 *10^{-9} }          

=>   E_c =  3.614 *10^{-19} \  J          

Generally the  number of photons emitted by the Sun in a second is mathematically represented as

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2 years ago
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2 years ago
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Answer is: <span>1/4 its old kinetic energy .
</span>V₁ = 10 m/s.
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m₁ = m₂ = m.
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m - mass of semi-truck.
E - kinetic energy of semi-truck.

5 0
3 years ago
Read 2 more answers
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