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kumpel [21]
3 years ago
6

A machine is supplied energy at a rate of 4000 watts and does useful work at a rate of 3761 watts what

Physics
1 answer:
Sergeu [11.5K]3 years ago
8 0
The efficiency of the machine is

                                 output energy  /  input energy

                             =  output power  /  input power

                             =     3,761 watts  /  4,000 watts 

                             =            (3,761  /  4,000)

                             =                0.94025

                             =                  94.025 %
                
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An archer shoots an arrow at a 74.0 m distant target, the bull's-eye of which is at same height as the release height of the arr
GuDViN [60]

A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °

B. The arrow will go over the branch.

<h3>A. How to determine the angle</h3>
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  • Initial velocity (u) = 33 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Angle (θ) = ?

R = u²Sine(2θ) / g

74 = 33² × Sine (2θ) / 9.8

Cross multiply

74 × 9.8 = 33² × Sine (2θ)

725.2 = 1098 × Sine (2θ)

Divide both sides by 1098

Sine (2θ) = 725.2 / 1098

Sine (2θ) = 0.6605

Take the inverse of sine

2θ = Sine⁻¹ 0.6605

2θ = 41.3

Divide both sides by 2

θ = 41.3 / 2

θ = 20.7 °

<h3>B. How to determine if the arrow will go over or under the branch</h3>

To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:

  • Initial velocity (u) = 33 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Angle (θ) = 20.7 °
  • Maximum height (H) = ?

H = u²Sine²θ / 2g

H = [33² × (Sine 20.7)²] / (2 ×9.8)

H = 6.94 m

Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).

Therefore, we can conclude that the arrow will go over the branch

Learn more about projectile motion:

brainly.com/question/20326485

#SPJ1

3 0
2 years ago
How much energy does the electron have initially in the n=4 excited state?what is the change in energy if the electron from part
Ierofanga [76]
What it looks to be that you found in A was the "initial"...b/c the question asks: 
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>

<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>

The answer comes to: <span>−1.36×10^−19 J</span>


You use the same equation for the second part as for part a. 
<span>just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. </span>
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Answer:

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