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LUCKY_DIMON [66]
3 years ago
8

1. An airow is shot horizontally from a crossbow 1.5 m above the ground. The initial velocity is 45

Physics
1 answer:
Amanda [17]3 years ago
4 0

Answer:

Time taken by the arrow to travel along to hit the ground is 0.55 seconds.

Explanation:

The only "force" acting on the "crossbow" to cause it to "hit" the ground is "gravity". There is no initial velocity downward when it shoot.

d=v_{i} t+\frac{1}{2} t^{2}

d = the displacement of the object  

t = the time for which the object moved  

a = acceleration of the object  

v_i = the initial velocity of the object

Given values

d = 1.5 m

t = unknown

a=g=9.8 \mathrm{m} / \mathrm{s}^{2}

\mathrm{V}_{\mathrm{i}}=0 \mathrm{m} / \mathrm{s}

1.5 \mathrm{m}=0(\mathrm{t})+\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \mathrm{t}^{2}

1.5=0+4.9 \mathrm{t}^{2}

\mathrm{t}^{2}=\frac{1.5}{4.9}

t^{2}=0.306 \mathrm{s}

Square root both sides

t=\sqrt{0.306}

t = 0.55 s

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Explanation:

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The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

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\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

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\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}.

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\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

At an efficiency of 0.25, the actual amount of energy required to raise this weight to that height would be:

\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0
1 year ago
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