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3 years ago

GETTING TIMED PLS HELP! Do the runners in the picture above represent kinetic or potential energy? you need to explain why.

Physics

1 answer:

prohojiy [21]3 years ago

8 0

Answer:

<em>They represent kinetic energy</em>

Explanation:

<u>Kinetic Energy </u>

A body can do work due to some of its attributes or states. For example, its mass can do work if used to provide energy, if the object is at a certain height respect to some reference level, it can do work when going downwards (potential energy), if the object moves at a certain speed, it can do work when transferring part of its speed to other objects. It's called kinetic energy and is given by

$\displaystyle K=\frac{mv^2}{2}$

Both runners are moving in a horizontal path, thus they have kinetic energy, given by the above equation. If they could jump below ground level, then they will also have potential energy

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A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

brainly.com/question/9805263

#LearnwithBrainly

7 0

3 years ago

What is my displacement if i walked 200 meters west then turned around and walked 150 meters east

Alinara [238K]

You're walking in one direction, and then the exact opposite of that direction, so you simply have to subtract the two distances.
200-150=50
You're 50 meters west of where you originally started.
You're west because 200 meters west is greater than 150 meters east. If the distance walked east was greater than the distance walked west, you would've been east of your starting position.

6 0

3 years ago

How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'

Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity $(K.E.=\frac{1}{2}mv^2)$ . If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height $(P.E.=mgh)$ . If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.

4 0

3 years ago

A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

Helga [31]

Answer:

2.59 m

Explanation:

Coordinates of origin = (0, 0)

Coordinates of Point p where the fly reach = (2.3 m, 1.2 m)

Use the distance formula of coordinates to find the distance between the origin and the point P.

$d=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$

$d=\sqrt{\left ( 2.3- 0 \right )^{2}+\left ( 1.2-0 \right )^{2}}$

d = 2.59 m

Thus, the distance between the origin and the point P is 2.59 m.

5 0

3 years ago

A friend of yours who has not taken astronomy sees a meteor shower (she calls it a bunch of shooting stars). The next day she co

Oxana [17]

Tell her meteors and stars aren't related a stars life time is extremely long and meteors are just rocks floating through space when they fly be the earth they can appear like falling stars but they aren't her fave constellation wont be going away any time soon

3 0

3 years ago