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Misha Larkins [42]
2 years ago
5

Who will it benefit if they do butterfly groin stretch​

Physics
2 answers:
Natali5045456 [20]2 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]2 years ago
3 0

Answer:

ummmmm mmmmm butterflies :)

Explanation:

You might be interested in
How do scientist study the earth through plate tectonic plates and boundaries/
Serggg [28]

Answer:

By plotting the locations of earthquakes

Explanation:

When plotting the locations of earthquakes, scientists have been able to locate plate boundaries and also be able to determine plate characteristics and predict the movement of plates

5 0
3 years ago
Lab: Energy Transfer Instructions Click the links to open the resources below. These resources will help you complete the assign
svetoff [14.1K]

Answer:

Give me a 5 star

Explanation:

Because im cool

6 0
2 years ago
Read 2 more answers
A girl delivering newspapers covers her route by traveling 5.00 blocks west, 5.00 blocks north, and then 7.00 blocks east. What
FromTheMoon [43]

Answer:

P = 2i + 5j

Therefore she is 2 blocks east and 5 blocks north.

Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks

Angle = taninverse (5/2)

Angle = 68.2°

Explanation:

Given:

Let west be negative and east be positive x axis.

Let north be the positive y axis.

5.00blocks west = -5.00 i

5.00 blocks north = 5.00 j

7.00 blocks east = 7.00i

Addition of the vector form of hee position is;

P = -5i +7i -5j

P = 2i + 5j

Therefore she is 2 blocks east and 5 blocks north.

Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks

Angle = taninverse (5/2)

Angle = 68.2°

3 0
2 years ago
A car of mass 150 kg is riding down at constant velocity. What is the normal force?
xz_007 [3.2K]

Answer:

1500N

Explanation:

Normal force = mg - F sin theta

constant velocity means acceleration = 0

F= ma = 150× 0 = 0N

thus;

normal force = mg = 150 × 10 = 1500N

4 0
2 years ago
Read 2 more answers
A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is
tiny-mole [99]

Answer:

F=mg(sin(\theta )-0.25 cos(\theta ))

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\

Using value of N from equation β in α we get value of force as

F=mg(sin(\theta )-\mu cos(\theta ))

Applying values we get

F=mg(sin(\theta )-0.25 cos(\theta ))

8 0
2 years ago
Read 2 more answers
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