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
Answer:
4.08 s
Explanation:
Let the passenger took "t" time to catch the train
so in this case the total distance moved by the train + 5 m = total distance moved by the passenger
so we will have
distance moved by train is given as
also the distance moved by passenger
so we will have
t = 4.08 s
Answer:
Yes, the energy is not simply the sum of the individual binding energies at each site, it is the product of energy at each binding site of hemoglobin.
Explanation:
Myoglobin and hemoglobin are two different cells. Myoglobin binds only one oxygen while the hemoglobin has the ability to binds four oxygen atoms at its four sides. Myoglobin present in muscle tissue only while hemoglobin is present in the whole body. Oxyhemoglobin is formed when oxygen binds with hemoglobin cell. This oxygen is take to all cells and energy is released due to the breakdown of glucose molecules with this oxygen.
4200 N is the tension in the cable that pulls the elevator upwards.
The correct option is A.
<h3>What does tension ?</h3>
Tension is the force that is sent through a rope, thread, or wire whenever two opposing forces pull on it. Along the whole length of the wire, the tensile stress pulls equally on all objects at the ends. Every physical object that comes into contact with that other one exerts force on it.
<h3>Briefing:</h3>
We employ the following formula to determine the cable's tension.
Formula:
T = mg+ma............ Equation 1
Where:
T is the cable's tension.
M = Mass of the elevator and the Joey
Accelerating with a
g = Gravitational acceleration
Considering the query,
Given:
m = (300+60) = 360 kg
a = 2 m/s²
g = 9.8 m/s²
Substitute these values into equation 2
T = (360×9.8)+(360×2)
T = 3528+720
T = 4248 N
T ≈ 4200 to the nearest hundred.
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The change in potential energy when the block falls to ground is -480J.
The maximum change in kinetic energy of the ball is 480 J.
The initial kinetic energy of the ball is 0 J.
The final kinetic energy of the ball is 0.148J.
The initial potential energy of the ball is 0.187 J.
The final potential energy of the ball is 0 J.
The work done by the air resistance is 0.039 J.
<h3>Change in potential energy when the block falls to ground</h3>
ΔP.E = -mgh
ΔP.E = -Wh
ΔP.E = - 40 x 12
ΔP.E = -480 J
<h3>Maximum change in kinetic energy of the ball</h3>
ΔK.E = - ΔP.E
ΔK.E = - (-480 J)
ΔK.E = 480 J
<h3>Initial kinetic energy of the ball</h3>
K.Ei = 0.5mv²
where;
- v is zero since it is initially at rest
K.Ei = 0.5m(0) = 0
<h3>Final kinetic energy</h3>
K.Ef = 0.5mv²
K.Ef = 0.5(0.0091)(5.7)²
K.Ef = 0.148 J
<h3>Initial potential energy of the ball</h3>
P.Ei = mghi
P.Ei = 0.0091 x 9.8 x 2.1
P.Ei = 0.187 J
<h3>Final potential energy</h3>
P.Ef = mghf
P.Ef = 0.0091 x 9.8 x 0
P.Ef = 0
<h3>Work done by the air resistance</h3>
W = ΔE
W = P.E - K.E
W = 0.187 J - 0.148 J
W = 0.039 J
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