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zvonat [6]
3 years ago
12

The power of a motor is 60kW. At what speed can it raise a load of 50000

Physics
1 answer:
ElenaW [278]3 years ago
8 0

Well right off the bat, you've handed us a serious problem.  You left the unit off of the 50,000, so we don't know the single most important thing that we need to know about the load.

Is the load 50,000 kilograms ? 50,000 Newtons ? 50,000 pounds ?  Each of these will have a different answer.

Since you didn't specify the unit, I can make it anything I want, and I can pick the unit to make the problem easy to solve.  

So I'm going to say that the load weighs 50,000 Newtons, and now I shall proceed to solve the problem that I have invented.

If you're using a motor that's marked "60 kW", that number is the maximum safe power the motor can deliver without overheating and breaking down.  In order to raise our 50,000N load as fast as possible, we'll run the motor at its maximum rated power of 60 kW.  That means it'll do 60,000 Joules of work for us every second.

Power = 60,000 Joules/second

Power = 60,000 Newton-meters/second

60,000 Newton-meters/sec = (50,000 Newton) x S meters/sec.

Divide each side by 50,000 Newtons:

S = (60,000 Newton-m/sec) / (50,000 Newton)

S = (60,000/50,000) meter/sec

<em>Speed = 1.2 meters/sec </em>

Our 60 kW motor can raise the load at the speed of 1.2 m/s or any slower, but no faster than that.

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An object has a mechanical energy of 1575 J and a potential energy of 1265 J.
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6.  

A. 1575 - 1265 = 310J

B.  KE 1/2 MV^2

   V=√2·KE/M = √2(310)/12   V = 7.2mls

C.  PE = 1265 = mgh

h= 1265/mg = 1265/(12)(92)    h= 10.8m    

7.  

A.  KE  = 1/2 mv^2  0.5(5)(12)^2    KE = 360J

B.  PE = mgh = (5)(9.8)(2.6)  PE = 127.4J

C.  ME = KE + PE = 360 + 127.4   ME = 487.4J

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3 years ago
When a nerve cell fires, charge is transferred across the cell membrane to change the cell's potential from negative to positive
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Explanation:

The given values are:

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q = 9.2 pC

Time,

t = 0.52ms

The equivalent circuit of the cell surface is provided by:

⇒  i_{avg}=\frac{charge}{t}

Or,

⇒  i_{avg}=\frac{q}{t}

On substituting the given values, we get

⇒         =\frac{9.2\times 10^{-12}}{0.52\times 10^{-3}}

⇒         =17.69^{-9}

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4 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
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a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

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2 years ago
Forces in a Three-Charge System Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separ
Dmitriy789 [7]

Answer:

Explanation:

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Q2=32.5nC. Positive charge

Q3=55nC. Positive charge

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Q2 is at the origin i.e at 0m

Q3 is at between Q1 and Q2 at -1.085m on the x-axis

It shows that,

Q1 is at -1.085+1.68 =0.595m from Q3

Also, Q2 is at 1.085m from Q3.

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We need to find the net force on Q3

Then we need F13 and F23

Firstly F13

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There will be attraction i.e, Q3 will move to the negative direction of the x axis, then F13 will be in negative direction

So,

F13=kQ1Q3/r²

F13=9E9×17.5E-9×55E-9/0.595²

F13=2.45×10^-5N

In vector form

F13=—2.45×10^-5N i

Now, we need F23,

This will the force of repulsion because they are both positive charge, the the charge Q3 will move to the negative direction of x axis, since Q2 is at the origin and Q3 is at negative x axis. So, F23 will be negative

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F23=9E9×32.5E-9×55E-9/1.085²

F23=1.367×10^-5N

In vector form

F23=—1.367×10^-5N i

Then the net force is given as

Fnet = F13+F23

Fnet=—2.45×10^-5Ni—1.367×10^-5Ni

Fnet=—3.82×10^-5N i

Magnitude for the Fnet is

Fnet=3.82×10^-5N.

And the direction

θ= arctan(y/x).

y=0 and x=3.82×10-5

θ= arctan(0/-3.82E-5)

θ=arctan(-0)

θ= 0. in the negative direction, i.e 180°.

3 0
3 years ago
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