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vlabodo [156]
3 years ago
12

The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply

upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 61° above the horizontal. With this launch angle, a skier attains a height of 16 m above the end of the ramp. What is the skier's launch speed?

Physics
1 answer:
kykrilka [37]3 years ago
4 0

Answer: Got It!

<em>Explanation:  </em>let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

0  0

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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat
Sever21 [200]

Answer:

Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, {\rm J}):

\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}.

Height of the weight (should be in meters, {\rm m}):

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

Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\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.

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1 year ago
What is the kinetic energy that the device contains immediately before it hits the ground? [KE = /2*m*v-]
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Answer:

Kinetic energy : "It is the energy possessed by body by virtue of its motion" .

Explanation:

The mechanical energy is the energy that a body has due to its motion or position .

It is of two types :

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It is given by P.E=mgh

Kinetic energy

"It is the energy that a body has due to its motion" .

The formula is : K.E=1/2mv².

For example

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