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Serga [27]
3 years ago
8

What is the relationship between Potential and Kinetic Energy

Physics
1 answer:
Wittaler [7]3 years ago
3 0
Potential energy is to stored with in an object due to position and arrangement. Kinetic energy is to energy with in an object due its to movement with their motion.

It will help you.

-Charlie

Have a great day
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What is unique regarding the abdominal muscle when is compared to other muscle in the body?
marissa [1.9K]
Answer:a is the answer
8 0
2 years ago
At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon
ArbitrLikvidat [17]

Answer:

The  pressure at point 2 is P_2  = 254.01 kPa

Explanation:

From the question we are told that

   The speed at point 1  is  v_1  =  3.57 \ m/s

   The  gauge pressure at point 1  is  P_1  =  68.7kPa =  68.7*10^{3}\  Pa

    The density of water is  \rho  = 1000 \ kg/m^3

Let the  height at point 1 be  h_1 then the height at point two will be

      h_2  =  h_1  -  18.5

Let the  diameter at point 1 be  d_1 then the diameter at point two will be

      d_2  =  2 * d_1

Now the continuity equation is mathematically represented as  

         A_1 v_1  =  A_2 v_2

Here A_1 , A_2  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e A=  \frac{\pi d^2}{4}]

   which can represent as

             A \ \  \alpha \ \  d^2

=>         A = c   d^2

where c is a constant

  so      \frac{A_1}{d_1^2}  =  \frac{A_2}{d_2^2}

=>          \frac{A_1}{d_1^2}  =  \frac{A_2}{4d_1^2}

=>        A_2  =  4 A_1

Now from the continuity equation

        A_1  v_1  =  4 A_1 v_2

=>     v_2  =  \frac{v_1}{4}

=>     v_2  =  \frac{3.57}{4}

       v_2  =  0.893 \  m/s

Generally the Bernoulli equation is mathematically represented as

       P_1 + \frac{1}{2}  \rho v_1^2 +  \rho *  g * h_1  =  P_2 + \frac{1}{2}  \rho v_2^2 +  \rho *  g * h_2

So  

         P_2  =  \rho  * g  (h_1 -h_2 )+P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )  

=>    P_2  =  \rho  * g  (h_1 -(h_1 -18.3)  + P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )

substituting values

        P_2  =  1000  * 9.8  (18.3) )+ 68.7*10^{3}  +  \frac{1}{2}  *  1000 ((3.57)^2 -0.893 ^2 )

       P_2  = 254.01 kPa

 

8 0
3 years ago
a student measures the speed of yellow light in water to be 2.00 x 10(to the 8th power) meters per second, 1.87 x 10(to the 8th
mamaluj [8]
<span>3) Neither precise or accurate. This is because of the deviation between the measurements, they vary and are not within a good range. And they are not close to the accepted value. In order to be precise the measurements have to be relatively close to each other, and to be accurate they have to be close to the accepted value.</span>
8 0
3 years ago
HELP PLEASE <br> Find the net force necessary for a 15 kg object to accelerate at 20 m/s/s.
tensa zangetsu [6.8K]

Answer:

<h3>The answer is 300 N</h3>

Explanation:

The force acting on an object given the mass and acceleration we use the formula

<h3>force = mass × acceleration</h3>

We have

force = 15 × 20

We have the final answer as

<h3>300 N</h3>

Hope this helps you

3 0
3 years ago
Next
natta225 [31]

Answer:

maybe it is the first one

7 0
3 years ago
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