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Igoryamba
2 years ago
12

The best way to prevent heat exhaustion while exercising is to

Physics
1 answer:
algol132 years ago
8 0
Take breaks and drink plenty of water not to much because it will make you sick sorry apparently pu ke is a bad word.
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A spacecraft of mass 1500 kg orbits the earth at an altitude of approximately 450 km above the surface of the earth. Assuming a
PSYCHO15rus [73]

Answer:

1.28 x 10^4 N

Explanation:

m = 1500 kg, h = 450 km, radius of earth, R = 6400 km

Let the acceleration due to gravity at this height is g'

g' / g = {R / (R + h)}^2

g' / g = {6400 /  (6850)}^2

g' = 8.55 m/s^2

The force between the spacecraft and teh earth is teh weight of teh spacecraft

W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N

8 0
3 years ago
Precautions to be taken for the displacement method
finlep [7]

Volume by Displacement. The displacement method (submersion, or dunking method) can be used to accurately measure the volume of the human body and other oddly shaped objects by measuring the volume of fluid displaced when the object is submerged.

The following precautions should be taken very observantly:-

The line of sight must be perpendicular to measuring scale to avoid parallax error. Formation of bubbles inside the cylinder should be completely avoided. Any bubbles within leads to wrong measurements.

8 0
3 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
Help please with number 7 !!!!!!! Plz ahhhhhh
baherus [9]

it is the Atlantic ocean

5 0
2 years ago
Compare and contrast static, sliding, and rolling friction.
Mama L [17]
Static friction - the friction that occurs when a force is being applied on a still object. 

Sliding friction - the friction that occurs upon a moving object; it is always lower than the static friction of the same object.

Rolling friction - the friction on a rolling object; it is actually a form of static friction with the contact point of the wheel with the surface.
4 0
3 years ago
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