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iVinArrow [24]
2 years ago
14

A car starts from rest with an acceleration of 5 ft/s. What is its velocity after it has gone 600 ft?

Physics
1 answer:
Soloha48 [4]2 years ago
3 0

Answer:

First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters

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Under state law, what is the blood-alcohol limit for legally operating a motor vehicle?
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Answer:

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Explanation:

The legal standard for drunkenness across the United States ranges from 0.10 to 0.08. If a person's BAC measures 0.08, it means that there are 0.08 grams (i.e., 80 mg) of alcohol per 100 ml of blood. The American Medical Association says that a person can become impaired when the blood alcohol level hits 0.05.

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3 years ago
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Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr
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Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

W = dP + dK\\

Where

dP = Change in Potential Energy

dK = Change in Kinetic Energy

Change in Potential Energy

P_{i} = mgh_{i}\\  P_{f} = mgh_{f}

Where

P_{i} = Initial Potential Energy

P_{f} = Final Potential Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

h_{i} = Initial Height = 0

h_{f} = Final Height = 30 m

Inputting the values we get the answer for dP

dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000

Change in Kinetic Energy

K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2

Where

K_{i} = Initial Kinetic Energy

K_{f} = Final Kinetic Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

v_{i} = Initial Velocity = 0 m/2

v_{f} = Final Velocity = 4 m/s

Inputting the values we get the answer for dK

dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000

Total Work

W = dP + dK\\

Inputting the values

W = 2943000 + 80000

W = 3,023,000

a) Finding the useful Power Output

P = \frac{W}{t}

Where

P = Power Output

W = Work Done = 3,023,000J

t = Time = 12s

Inputting the values

P = \frac{3,023,000}{12}\\ P = 251,916.667

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

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Answer:

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Explanation:

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