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mars1129 [50]
3 years ago
6

A ski tow operates on a slope of angle 14.0^\circ of length 250 m. The rope moves at a speed of 13.0 km/h and provides power for

54 riders at one time, with an average mass per rider of 75.0 kg.
Estimate the power required to operate the tow. ?
Physics
1 answer:
mina [271]3 years ago
7 0

Answer:

The power required to operate the tow is 34.75 kW.

Explanation:

Given that,

A ski tow operates on a slope of angle, \theta=14^{\circ}

Speed of the rope, v = 13 km/h = 3.62 m/s

Number of riders, n = 54

The average mass per rider of 75.0 kg

We need to find the power required to operate the tow. We know that the rate at which the work is done is called power operated by an object.

P=\dfrac{W}{t}\\\\or\\\\P=F\times v

Here,

P=54mg\times v\sin\theta\\\\P=54\times 75\times 9.8\times 3.62\times \sin(14)\\\\P=34758.8\ W

or

P = 34.75 kW

So, the power required to operate the tow is 34.75 kW. Hence, this is the required solution.

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You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it
VladimirAG [237]

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg  = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

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mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd  + u²

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v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

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If 31.25
valentina_108 [34]
<span>1 C = 6.24150965(16)×10^18 electrons

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Answer:

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2 years ago
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The radius of a dime is approximately r_{n2} = 9\cdot 10^{-3}m: if we assume that the radius of the nucleus is exactly this value, then we  can find how far is the electron by using the proportion
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from which we find
r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.
6 0
3 years ago
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