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mr Goodwill [35]
3 years ago
7

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is

the maximum vertical height the person in the cart can reach?
Physics
1 answer:
HACTEHA [7]3 years ago
5 0

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

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Question:

The operations manager for a well-drilling company must recommend whether to build a new facility, expand his existing one, or do nothing. He estimates that long-run profits (in $000) will vary with the amount of precipitation (rainfall) as follows:

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B. $170,000

C. $285,000

D. $305,000

E. $475,000

Answer:

D. $170,000

Explanation:

The expected long run profits are  for

                        Low                Normal                 High

Do nothing    -100*0.3                100*0.2                   300*0.5 = 140

Expand           350*0.3                500*0.2                  200*0.5 = 305

Build new       750*0.3                300*0.2                  0*0.5    = 285

Therefore the expected long run profits are

$140,000

$305,000

$285,000

Based on his selected option being either to build new or to expand, the most profitable option is to expand

=$305,000

EVPI = EPPI-EMV =$170,000

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