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3 years ago

Who will it benefit if they do butterfly groin stretch

Physics

2 answers:

Natali5045456 [20]3 years ago

4 0

Answer:

It is a great stretch for athletes who play field or court sports, runners, and anyone who has tight hips or a history of groin injuries.

Its a great flexibility exercise

Explanation:

Hope that helps

igor_vitrenko [27]3 years ago

3 0

Answer:

ummmmm mmmmm butterflies :)

Explanation:

You might be interested in

What can be added to an atom to cause a nonvalence electron in the atom to temporarily become a valence electron

Inessa05 [86]

Answer:

providing energy to an atom can allow the electron in its non valence shell to obtain energy and move to a higher energy orbital and act as a valence electron.

Explanation:

8 0

3 years ago

Read 2 more answers

A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40

Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is $\frac{1}{2}mv^2$

So total energy at the top of the hill $=mgh+0=mgh$

At the bottom of the hill kinetic energy is equal to $\frac{1}{2}mv^2$ and potential energy will be 0

So total energy at the bottom of the hill is equal to $0+\frac{1}{2}mv^2$

Form energy conservation $mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$ , for v = 20 m/sec

$20=\sqrt{2\times 9.8\times h}$

Squaring both side

$19.6h=400$

h = 20.408 m

Now if velocity is 0 m/sec

$40=\sqrt{2gh}$

$19.6h=1600$

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

5 0

3 years ago

A 970-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

vlabodo [156]

Answer:

22.73 m/s or 81.72 kph

Explanation

We can find the combined mass of both cars as

970 kg + 2300 kg = 3270 kg.

Then the normal force of the cars can be calculated as

F(n)= mg

Where g is acceleration due to gravity 9.8m/s^2

3270 kg ×9.8 = 32046 kg*m/s^2.

coefficient of kinetic friction between tires and road to be 0.80 × F(n)

Then the frictional force can be calculated as

= (32046kg*m/s^2 × 0.80 )

= 25636.8 kg*m/s^2

We can now calculate the work done that was used stopping the cars as

Frictional force × distance

(25636.8 kg*m/s^2 ) × 2.9m= 74346.72kg*m^2/s^2

From kinetic energy formula, the combined velocity of the car can be determined

E=0.5 M V²

√(2E/M) = V

√(2*74346.72kg*m^2/s^2 / 3270 kg) = V

V= √ (45.472)

V=6.743293m/s

the momentum of both cars can be determined as

6.743293m/s * 3270 kg

= 22050.57kg*m/s

Now the final momentum of both cars must be equal to the the momentum of

the sports car just prior to the collision. Therefore, the speed of the sports car at impact.

=(22050.57 kg*m/s) / 970 kg = 22.73 m/s

We can convert that to km/h.

22.73 m/s * 3600 s/h / 1000 m/km = 81.72 kph

7 0

3 years ago

A car at the top of a ramp starts from rest and rolls to the bottom of the ramp, achieving a certain final speed. If you instead

zimovet [89]

Answer:

It must be 4 times high.

Explanation:

Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.
This means, that at any time, the following must be true:
ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)

⇒ $m*g*h = \frac{1}{2} * m*v^{2}$