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2 years ago

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A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat

three people. The lift is operating a a steady speed of 10 km/h. Neglecting friction and air drag and assuming the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also estimate the power required to accelerate this ski lift in 5 s to its operating speed when it is first turned on. (steady power = 68.1 kW; start-up = 43.7 kW)

1 answer:

myrzilka [38]2 years ago

4 0

Answer:

P = 68.125 kW

P startup = 43.05 kW

Explanation:

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Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The

Eddi Din [679]

Answer:

(a) Steel rod: $1.1 * 10^{-4}$

Copper rod: $1.88 * 10^{-4}$

(b) Steel rod: $8.3 * 10^{-5} m$

Copper rod: $1.41 * 10^{-4} m$

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = $\frac{1}{Y} * \frac{F}{A}$

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y = 200 000 000 000 $N/m^{2}$

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

∴ $Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}$

For the copper rod:

Y = 120 000 000 000 N/m²

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

$Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain = = 1.88 * 10^{-4}$

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = $1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m$

For the copper rod:

Elongation = $1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m$

6 0

2 years ago

A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an angle of 35°

spin [16.1K]

<em>Answer</em>

Force = 170 N

<em>Explanation</em>

First find the distance (d) travelled by the bulldozer.

Sin 35 = 15/d

d = 15/(sin 35)

= 26.15m

Now;

work done = force × distance.

4500 J = force × 26.15

dividing both sides by 26.15,

Force = 4500/26.15

= 172.07 N

Answer to two significant figures = 170 N

3 0

2 years ago

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

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2 years ago

In which of the following examples does the object have both kinetic and potential energy? Select all that apply.

notsponge [240]

I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic

6 0

2 years ago

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A boy throws a water ballon such that it hits his sister standing 10m away. The boy threw the water balloon at an angle of 35 de

Tresset [83]

Answer: 7.734 m/s

Explanation:

We have the following data:

$\theta=35\°$ The angle at which the water ballon was thrown

$x=10 m$ The horizontal distance of the water ballon

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find the initial velocity $V_{o}$ at which the water ballon was thrown, and we can find it by the following equation:

$x=V_{o}cos \theta T$ (1)

Where $T=2t$ is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time $T$ it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (2)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (3)

Remembering $T=2t$ :

$T=2\frac{-V_{o}}{g}$ (4)

Substituting (4) in (1):

$x=V_{o}cos \theta (2\frac{-V_{o}}{g})$ (5)

Isolating $V_{o}$ :

$V_{o}=\sqrt{\frac{x g}{-2 cos \theta}}$ (6)

$V_{o}=\sqrt{\frac{(10 m)(9.8 m/s^{2})}{-2 cos(35\°)}}$ (7)

Finally:

$V_{o}=7.734 m/s$

4 0

2 years ago