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Cloud [144]
3 years ago
14

In 1.71 minutes, a ski lift raises four skiers at constant speed to a height of 148 m. The average mass of each skier is 62.9 kg

. What is the average power provided by the tension in the cable pulling the lift?
Physics
1 answer:
Vitek1552 [10]3 years ago
6 0

Answer:

3560.36 Watts

Explanation:

Power, P=\frac {nΔW}{Δt) where P is power, n is the number of skiers, t is time in seconds and Δt is change in time, ΔW is given by mgh where m is mass, g is gravitational constant, h is height

Substituting n for 4 skiers, m for 62.9 Kg, g for 9.81, h for 148 m and t for 1.71*60=102.6 seconds

P=\frac {4*62.9*9.81*148}{1.71*60}=3560.360702  Watts

Average power is approximately 3560.36 Watts

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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
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Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
3 years ago
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