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Lynna [10]
3 years ago
5

How long does it take someone to run 8 miles if they are running at a speed of 7 km/hr?

Physics
1 answer:
ZanzabumX [31]3 years ago
7 0

Answer:

1.84 hours or 110.4 minutes

Explanation:

8 miles             1.61 km                 1 hr

------------    x   -----------------  x   -------------- = 1.84 hours or 110.4 minutes

                          1 mile                7  km    

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Most energy obtained from water is converted from _____.
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Potential energy behind dams
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3 years ago
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Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
3 years ago
A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic
zhannawk [14.2K]

Answer:

The force is  F  = 172 \ N

Explanation:

From the question we are told that

    The  mass of the block is  m_b  = 27.0 \ kg

     The  coefficient of  static friction is  \mu_s  =  0.65

     The coefficient of kinetic friction is  \mu_k  =  0.50

The  normal force acting on the block is  

      N  =  m *  g

substituting values

     N  =   27 *  9.8

     N  =   294.6  \  N

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

        F_f  =  \mu_s  *  N

substituting values

        F_f  =   0.65 *  264.6

        F_f  =   172 \ N

Now for this  block to move the force require is  equal to F_f i.e

       F= F_f

=>    F  = 172 \ N

       

   

5 0
3 years ago
How do you calculate resistance with coulombs, time and J's?
Mars2501 [29]
Hey user


The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C):

E(J) = V(V) ×<span> Q</span>(C)

So

joule = volt × coulomb

or

J = V × C

Example

What is the energy in joules that is consumed in an electrical circuit with voltage supply of 15V and charge flow of 4 coulombs?

E = 15V × 4C = 60J

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3 0
4 years ago
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