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Ludmilka [50]
3 years ago
12

A 45000 watt crane operating at full power lifts a 2100 kg object vertically for 17.4 seconds. How high has the crane lifted the

object?
Physics
1 answer:
LenKa [72]3 years ago
8 0

Answer:

Explanation:

We need the power equation here, which is:

Power = (F * Δx)/time

where F * Δx is the amount of work done.

F is a force which is measured in Newtons. We are given the mass of the object, but since we need a Force measure, we need the weight of the object:

F = 2100(9.0)

F = 21000 to the correct number of sig dig.

Now we can plug in the values we have and solve for the displacement, Δx:

45000=\frac{21000x}{17.4} and isolating x:

\frac{17.4(45000)}{21000}=x so

x = 37 m

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labwork [276]

Answer:

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Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At s
Tresset [83]

Answer:

n=0.03928 moles

Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles

Explanation:

The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs

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Pressure at sea level = P= 1 atm=1.0125*10^5 Pa

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Formula:

n=\frac{PV}{RT}\\n=\frac{(1.0125*10^5) *(1*10^{-3})}{(8.314)*310} \\n=0.03928 mol

Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles

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4 years ago
Brewed coffee is often too hot to drink right away. you can cool it with an ice cube, but this dilutes it. or you can buy a devi
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3 years ago
The 5-lb collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it s
Sholpan [36]

Answer:

Explanation:

Stiffness of spring k equal to 4 lb/ft.

Unstretched length of the spring L is equal to 0.5 feet.  

Weight of the collar W is 15lb

Radius of curvature of curve guide is 1 feet

length of vertical rod is 1.5 feet

Initial speed of collar when released from rest at A is 0 feet per seconds  

use the energy conservation equation

 P_A +K_A=P_B+K_B

Estimate the potential energy , component as position B as below

P_A=Wh_1+\frac{1}{2} ks^2_1\\\\=5\times(1.5+1)+\frac{1}{2} \times 4 \times(1.5+1-0.5)^2\\\\=20.5lb.ft

Estimate the kinetic energy , component as position A as below

K_A=\frac{1}{2} \frac{W}{g} V^2_1\\\\=\frac{1}{2} \frac{5}{32.2} \times0^2\\\\=0lb.ft

Estimate the kinetic energy , component as position B as below

K_A=\frac{1}{2} \frac{W}{g} V^2_2\\\\=\frac{1}{2} \frac{5}{32.2} \times V^2_2\\\\=00777V^2_2

Substitute 20.5lb- ft for P_A

0.5lb-ft for P_B

0lb -ft for K_A

0.0777V_2^2 for K_B

20.5+0=0.5+0.777V^2_2\\\\V^2_2=257.6\\\\V_2=16.05

= 16.05ft/sec

7 0
3 years ago
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