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pshichka [43]
3 years ago
15

PLEASE HURRY IM BEING TIMED

Physics
1 answer:
spin [16.1K]3 years ago
8 0
A compound machine is one that consists of a couple simple machines working together, like a wheelbarrow contains the lever, axle, and wheel.  While a simple machine would just be those individually, or anything else that can be used for applying a force.<span />
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Find the force exerted by each rope in the sign below
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Answer: Example 1: Consider a crate being pulled along a frictionless floor (while such a floor is very hard to find, this will still help us understand the concept and we can return to this situation later, after considering friction, and solve it more realistically).

Consider a crate being pulled along a horizontal, frictionless floor. A rope is tied around it and a man pulls on the rope with a force of T. T is the tension in the rope. What happens to the crate?

Before we can apply Newton's Second Law,

F = m a

we must find the net force -- the vector sum of all the forces -- acting on the object. In addition to the force T exerted by the rope, what other forces act on the object?

As discussed in class, in Mechanics, we can restrict our attention to "contact" forces and "gravity". That means gravity pulls down on this crate with a force equal to its weight, w. But the floor supports the crate. The floor responds by pushing up on the crate with a force we call the normal force. "Normal" means "perpendicular". We will call this force n; you may also encounter it labeled N or FN.

Explanation:

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3 years ago
Sixty identical drippers, each with a hole of diameter 1.00 mm, are used to water a yard. If the water in the main pipe of diame
VashaNatasha [74]

Answer:

a

   V =  5.30 *10^{-2} \ m^3

b

   v_1 = 0.3127 \ m/s

Explanation:

From the question we are told that

   The number of identical drippers is  n =  60

   The diameter of each hole in each dripper is  d =  1.0 \  mm =  0.001 \  m  

   The diameter of the main pipe is  d_m  =  2.5 \  cm  =  0.025 \  m

    The speed at which the water is flowing is  v  =  3.00 \  cm/s =  0.03 \  m/s

Generally the amount of water used in  one hour = 3600 seconds is mathematically represented as

          V =  A *  v  *  3600

Here A is the area of the main pipe with value

         A =  \pi  * \frac{d^2}{4}

=>       A = 3.142   * \frac{0.025^2}{4}

=>        A =  0.0004909 \  m^2

So  

=>   V =  0.0004909  *  0.03  *  3600

=>  V =  5.30 *10^{-2} \ m^3

Generally the area of the drippers is mathematically represented as

       A_1=  n  * \pi \frac{d^2}{4}

=>    A_1 =  60   * 3.142 *  \frac{0.001 ^2}{4}

=>    A_1 =  4.713 *10^{-5} \  m^2

Generally from continuity equation we have that  

         Av =  A_1 v_1

=>      0.0004909 *  0.03 =  4.713 *10^{-5} *  v_1

=>   v_1 = 0.3127 \ m/s

   

     

3 0
3 years ago
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