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MaRussiya [10]
3 years ago
15

Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find:

Time of flight? Horizontal distance traveled? Maximum height?
Physics
1 answer:
grandymaker [24]3 years ago
6 0

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, \theta=35^{\circ}

Time of flight :

t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s

Horizontal distance traveled  is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

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Using 15 percent as machine energy efficiency, what is the actual work output if the total work input is 7500 kilojoules​?
g100num [7]

Answer:

The actual work output is 1,125 kilojoules.

Explanation:

Energy efficiency is defined as the efficient use of energy. An appliance, process, or facility is energy efficient when it consumes less than the average amount of energy to perform an activity. Then, energy efficiency is the ratio between the amount of energy used in an activity and the amount expected to be carried out.

Efficiency is calculated as:

efficiency = output / input

Where output is the amount of mechanical work (in watts) or energy consumed by the process (in joules), and input (input) is the amount of work or energy that is used as input to carry out the process.

In this case:

  • Efficiency always has a value between 0 and 1. In this case, efficiency=0.15
  • output=?
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Replacing:

0.15=output/7500 kilojoules

Solving:

Output=0.15* 7500 kilojoules

output=1,125 kilojoules

<u><em>The actual work output is 1,125 kilojoules.</em></u>

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Short focal length gives a deeper depth of the present field in addition to the wide angle of view. They capture maximally that lies in their front of the camera to the infinity.

A lens with large focal length captures a narrow field band as compared to the one with short focal length.

The lower focal length shows the short depth of focus leading to less magnification.

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