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3 years ago

Racewalking involves more impact than running. true false

Physics

2 answers:

ahrayia [7]3 years ago

6 0

Answer:

Given statement is false.

Explanation:

We have been given a statement that racewalking involves more impact than running.

The given statement is false.

Running involves more effort that is number of calories burn will be more. So, running involves more impact.

Running is good for cardiovascular health that is heart health.

natta225 [31]3 years ago

5 0

I think the statement is false. Racewalking involves less impact than running. It is a long-distance discipline within the sport of athletics. Although it is a foot race, it is different from running in that one foot must appear to be in contact with the ground.

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3 years ago

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

Horizontally:

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

Vertically:

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :

$T=2t=2(5.61 s)=11.22 s$ (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

$x=V_{ox}T$ (10)

Substituting (2) and (9) in (10):

$x=(18.811 m/s)(11.22 s)$ (11)

Finally:

$x=211.059 m$

8 0

3 years ago