Answer: 37.5 kg in 3 s.f.
Explanation:
Answer:
all of those are pisitions
Explanation:
Answer:
The value is 
Explanation:
From the question we are told that
The power rating of the stove is 
The duration of its use everyday is 
The rating of the light bulbs is 
The number is 
The power rating of the total bulb is 
The duration of its use everyday is 
The power rating of miscellaneous appliance 
The duration of its use everyday is 
The power rating of hot water 
The duration of its use everyday is 
Generally the total electrical energy used in 1 month is mathematically represented as
=> 
=> 
=> 
Generally the monthly electricity bill is mathematically represented as
=>
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
