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1 year ago

9

Calculate the potentia energy of a car with a mass of 3800kg that is on a hill 110 meters above sea level? USE 10 instead of 9.8

for the acceleration due to gravity

1 answer:

Arada [10]1 year ago

7 0

Potential energy (PE ) = m g h

Where:

m = mass = 3800 kg

g = acceleration due gravity = 10 m/s^2

h = heigth = 110 meters

Replacing:

PE = 3800 * 10 * 110 = 4,180,000 J

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Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

nalin [4]

Answer:

$13 W/m^2$

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

$I \propto \frac{1}{r^2}$

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

$\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2}$ (1)

where in this problem, we have:

$I_1 = 1300 W/m^2$ apparent brightness at a distance $r_1$ , where

$r_1 = 150$ million km

We want to estimate the apparent brightness at $r_2$ , where $r_2$ is ten times $r_1$ , so

$r_2 = 10 r_1$

Re-arranging eq.(1), we find $I_2$ :

$I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2$

5 0

3 years ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago

Any two application of gravity

Brrunno [24]

Answer:

Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.

Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.

The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

5 0

3 years ago

A little girl kicks a soccer ball. It goes 10 feet and comes back to her. How is this possible?

marta [7]

A classic puzzle...

She either kicked it at a wall <em>exactly</em><em /> 10 foot in front of her, where the ball rebounded off the wall.

Or, she kicked the ball straight up, vertically, at a <em>90 degree angle,</em> where due to the law of gravity, which states that anything that goes up must come down, when the soccer ball reaches exactly 10 feet, it falls back down.
(Note: This is nearly impossible to achieve -- exactly 10 feet.)

8 0

4 years ago

Rachel collected data on the speed of sound waves passing through different media.

Helga [31]

The conclusion is, medium Q is most likely a solid because solids have the highest density and sound waves travel fastest in high density media.

<h3>Effect of density on speed of sound</h3>

Sound wave is mechanical wave that requires material medium for its propagation.

A high dense medium, is a medium with closely packed molecules. Since sound wave requires material medium for its propagation, it will travel faster in a high dense medium than a less dense medium.

Thus, the speed of sound increases as the density of the medium increases.

<h3>Speed of sound in the different media</h3>

The conclusion that can be made from the speed of sound in the different media is "Medium Q is most likely a solid because solids have the highest density and sound waves travel fastest in high density media".

Learn more about effect of density on speed of sound here: brainly.com/question/3323620

7 0

2 years ago