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MrMuchimi
3 years ago
5

An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit

h an average mass of 70.0 kg and it takes 20.0 s to accomplish this task, determine the mass of the elevator.
Physics
1 answer:
mylen [45]3 years ago
4 0

Find how much work ∆<em>W</em> is done by the motor in lifting the elevator:

<em>P</em> = ∆<em>W</em> / ∆<em>t</em>

where

• <em>P</em> = 45.0 kW = power provided by the motor

• ∆<em>W</em> = work done

• ∆<em>t</em> = 20.0 s = duration of time

Solve for ∆<em>W</em> :

∆<em>W</em> = <em>P</em> ∆<em>t</em> = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is <em>M</em> = (<em>m</em> + 490.0) kg, where <em>m</em> is the mass of the elevator alone. Then

∆<em>W</em> = <em>M</em> <em>g h</em>

where

• <em>g</em> = 9.80 m/s² = acceleration due to gravity

• <em>h</em> = 35.0 m = distance covered by the elevator

Solve for <em>M</em>, then for <em>m</em> :

<em>M</em> = ∆<em>W</em> / (<em>g h</em>) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

<em>m</em> = <em>M</em> - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

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The ratio of the distance between Earth and Pluto and the distance between Earth and Mars = (5.2241 × 10⁹ km)/(220.58 × 10⁶ km) ≈ 23.683

It took 2006 to 2015 (9 years) to travel from Earth to Pluto, therefore, it can take approximately (9 years)/(23.683) ≈ 0.38 of a year which is ((9 years)/(23.683)) × 365.2422 ≈ 138.8 days for a spacecraft to travel from Earth to Mars

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3 years ago
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

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Explanation:

Given:

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  • speed of lagging car, u_{2}=35\ m.s^{-1}
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a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

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3 years ago
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3 years ago
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Put the value into the formula

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6 0
3 years ago
2500000000000 in scientific notation
Alex_Xolod [135]
Your answer is 2.5 x 10^(11).

Please take the time to rate and thank me if you feel I helped you! Thanks, it means a lot to me. :)
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3 years ago
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