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The ability of your joints and muscles to move in their full range of motion is called

Studentka2010 [4]

Answer:

Dynamic flexibility

Explanation:

Dynamic flexibility can be generally defined as the ability of the body muscles and joints to move in full range of motion. High flexibility in these joints and muscles leads to the decreasing pain and injury in different parts of the body.

Proper warm up exercises are needed to be carried out that involves both the combination of controlling movements and stretching of the body, and this directly enhances the dynamic flexibility of the body.

The athletes and sports persons possesses a good dynamic flexibility of their body as they carry our different types of body exercises.

6 0

3 years ago

If the speed of a car is 20m/s .How long does it take to cover a distance of 1km?

3241004551 [841]

Answer: 50 seconds

Explanation:

1km=1000m

Distance/speed=time

1000/20=50

50 seconds

7 0

3 years ago

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Principal of simple machine?

sergij07 [2.7K]

Answer:

if there is no friction in a simple machine, work output and work input are found equal in that machine

Explanation:

3 0

3 years ago

A car initially traveling at 17.1 mph comes to rest in 9.7s what was its acceleration in this time?

ra1l [238]

Answer:

$a=-.78m/s^2$

Explanation:

Δ $v=at$

Δ $v$ is the difference in velocity before and after a given time.
$a$ is the acceleration of the object during this time.
$t$ is time

$(v_f-v_i)=at$ is another way to write this equation.

The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δ $v=(v_f-v_i)$

$v_f-v_i=at\\\frac{at}{t}=\frac{v_f-v_i}{t}\\a=\frac{v_f-v_i}{t}$

I rearranged this equation to solve for $a$ , but this is a step that you don't need to take, it's just good to get in the habit of doing this.
Plug in the given values. Note that our final velocity is $0$ , because the car travels until at rest.

$a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}$

Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.

$a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}$

if you converted correctly, your answer for $v_f$ will be ≅ $7.6m/s$ .
Now divide. Notice that the units for acceleration are $m/s^2$ or meters per second, per second.

$a=\frac{-7.6m/s}{9.7s}\\a=-.78m/s^2$

Our final answer is negative because the car is slowing down. Do not square this answer as the square symbol only applies to the units, not the magnitude.

4 0

3 years ago

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An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector

Ghella [55]

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

$v_{x}$ = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

$t = \frac{X}{v_{x}}$

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel = 3.67 x 10⁻⁸ sec

$v_{y}$ = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = $v_{y}$ t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

7 0

3 years ago