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8_murik_8 [283]

2 years ago

11

the taipei 101 in taiwan is a 1667-foot tall, 101-story skyscraper. the skyscraper is the home of the world's fastest elevator.

the elevators transport visitors from the ground floor to the observation deck on the 89th floor at speeds up to 16.8 m/s. determine the power delivered by the motor lift the 10 passengers at this speed. the combined mass of the passengers and cabin is 1250 kg

1 answer:

BabaBlast [244]2 years ago

6 0

The power delivered by the motor lift in elevating the 10 passengers at the speed of 16.8 m/s is 205800 Watts

<h3>What is power? </h3>

Power is simply defined as the rate at which energy is consumed. It can be expressed mathematically as

Power (P) = Force (F) × velocity (v)

P = Fv

<h3>How to determine the force </h3>

Mass (m) = 1250 Kg
Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 1250 × 9.8

F = 12250 N

<h3>How to determine the power </h3>

Velocity (v) = 16.8 m/s
Force (F) = 12250 N

Power (P) =?

P = Fv

P = 12250 × 16.8

P = 205800 Watts

Learn more about power:

brainly.com/question/5684937

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Suppose that a spring with a larger spring constant was used in this same apparatus. If a given mass were rotated at the same ra

malfutka [58]

Answer:

The new period of rotation using the new spring would be less than the period of rotation using the original spring

Explanation:

Generally the period of rotation of the mass is mathematically represented as

$T = 2 \pi \sqrt{\frac{I}{k} }$

Here I is the moment of inertia of the mass about the rotation axis and k is the spring constant

Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser

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3 years ago

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound

Sophie [7]

Answer:

a. Wavelength = λ = 20 cm

b. Next distance of maximum intensity will be 40 cm

Explanation:

a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.

Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.

This we get the equation:

(nλ + λ/2) - nλ = 30-20

λ/2 = 10

λ = 20 cm

b. at what distance, sound intensity will be maximum again.

For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.

= 30 + λ/2

= 30 + 20/2

=30+10

=40 cm

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3 years ago

Why does a dropped feather hit the ground later than a rock dropped at the same time? *

MissTica

The air resistance on the feather would be the correct answer.

7 0

3 years ago

Read 2 more answers

A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the

kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

$E=\frac{kQ}{r^2}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$Q=6 \mu C = 6\cdot 10^{-6}C$ is the charge producing the field

r = 100 m is the distance from the charge at which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

$E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C$

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3 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

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3 years ago