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yawa3891 [41]
2 years ago
5

HELP ASAP REAL ANSWERS PLEASE You push a box across the floor with a force of 30 N. You push it 15 meters in 8 seconds. How much

work did you do? How much power did you use?
Physics
2 answers:
dalvyx [7]2 years ago
6 0

When you lift a box up off the floor, you must exert a force at least equal to the weight of the box. If you lift the box at constant speed, you "do work" on the box that would be equal to the force you exert (the weight of the box) times the height through which you lifted the box. If you exert a force of 50 Newtons on the box and lift it 1 meter high, then you did (50 N)x(1 m) or 50 N-m of work on the object. 1 Newton-meter is called a Joule (J), pronounced "jewel." In lifting the 50 N box upward 1 meter at constant speed, you would do 50 Joules of work on the box. That means your body used 50 J of energy to lift the box.

<span>and here is the Equation for Work: Work = Force x distance, or W = F d.</span>


Leya [2.2K]2 years ago
6 0

When you lift a box up off the floor, you must exert a force at least equal to the weight of the box. If you lift the box at constant speed, you "do work" on the box that would be equal to the force you exert (the weight of the box) times the height through which you lifted the box. If you exert a force of 50 Newtons on the box and lift it 1 meter high, then you did (50 N)x(1 m) or 50 N-m of work on the object. 1 Newton-meter is called a Joule (J), pronounced "jewel." In lifting the 50 N box upward 1 meter at constant speed, you would do 50 Joules of work on the box. That means your body used 50 J of energy to lift the box.

<span>and here is the Equation for Work: Work = Force x distance, or W = F d.</span>

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Wavelength = speed / frequency

(345 m/s) / (20,000 Hz) = 0.017 m

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Calculate the Potential Energy of an object that has a mass of 14-kg and is located at a height of 24-m?
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Answer:

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Explanation:

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Mass Center Determine the coordinates (x, y) of the center of mass of the area in blue in the figure below. Answers: x=(3)/(8)a
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Explanation:

The x and y coordinates of the center of mass are:

xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA

ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA

Assuming uniform density, the center of mass is also the center of area.

xcm = ∫ x dA / ∫ dA = ∫ x y dx / A

ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A

First, let's find the area:

A = ∫ y dx

A = ∫₀ᵃ (-h/a² x² + h) dx

A = -⅓ h/a² x³ + hx |₀ᵃ

A = -⅓ h/a² (a)³ + h(a)

A = ⅔ ha

Now, let's find the x coordinate of the center of mass:

xcm = ∫ x y dx / A

xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)

xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)

xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)

xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)

xcm = (¼ ha²) / (⅔ ha)

xcm = ⅜ a

Next, we find the y coordinate of the center of mass:

ycm = ∫ y² dx / A

ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)

ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ x⁵ − ⅔ h²/a² x³ + h² x) |₀ᵃ / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)

ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)

ycm = ⅖ h

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Soloha48 [4]
These are called neap tides!

Hope this helps :)
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