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2 years ago

Which of the following is an example of potential energy?

Physics

1 answer:

goblinko [34]2 years ago

3 0

Answer:

Explanation:

potential energy is energy due to position

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A 5.0 kg cannonball is fired horizontally at 68 m/s from a 15-m-high cliff. A strong tailwind exerts a constant 12 N horizontal

jeka57 [31]

Answer:

3.7 m

Explanation:

ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky

The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of

t = √ (2h/g) = √(2(15)/9.8) = 1.75 s

Without the tail wind, the ball travels horizontally

d = vt = 68(1.75) = 119 m

The tailwind exerts a constant acceleration on the ball of

a = F/m = 12/5.0 = 2.4 m/s²

The average horizontal velocity during the flight is

v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s

so the distance with tailwind is

d = v(avg)t = 70.1(1.75) = 122.675 m

The extra distance is 122.675 - 119 = 3.675 = 3.7 m

8 0

3 years ago

The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac

andrezito [222]

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

5 0

3 years ago

What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface

kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman $= \frac{95.08}{100 } = 0.9508$

The density of the woman is calculate as;

$Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water$

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0

3 years ago

How do you do this? Plz help answer

Contact [7]

Answer: Really

Explanation:

Just look it up for this page and maybe you will find an anwser sheet.

7 0

3 years ago

What variables affect density weight,conductivity,color,volume, or mass

tatiyna

Answer:

Density is affected by volume and mass.

Explanation:

Density is defined as the quantity of mass per unit of volume, or expressed mathematically, d = m/v.

6 0

3 years ago