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3 years ago

15

What happens to gas that is not used

1 answer:

kvv77 [185]3 years ago

3 0

the engine won't start or it sputters when it should be running perfectly. if the gasoline is old and stale, it will have lost a portion of its volatility. the lighter components of the gasoline (remember, gasoline is a mixture of different hydrocarbons) have probably evaporated off or disappeared.

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for a freely falling object dropped from rest, what is the acceleration at the end of the fifth second of fall?

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acceleration due to gravity is always 9.8 m/s/s (on earth)

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3 years ago

Were there experiences in his life that led to his interest in science and the study of the atom?

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The right answer is Democritus

Explanation:

He experienced bonding with his father which strongly resulted his interest in science and to the study of the atoms ;)

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2 years ago

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An 85 kg dog is sitting on a couch what is the weight of the dog

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3 years ago

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A telephone line hangs between two poles 14 m apart in the shape of the catenary , where and are measured in meters.

Masja [62]

Answer:

a) At x=14 the slope will be given by:

$\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right)$ .

b) Then, the angle between the line and the pole will be:

$\phi=\pi - \theta$

where $\theta$ is the angle between the tangent to the catenary and the x-axis.

Explanation:

The catenary has the following general form:

$y(x)==a\cosh \left({\frac {x-C_{1}}{a}}\right)+C_{2}$

a) The slope at any point will be given by the derivative of y.

$\frac{dy}{dx}(x)=a\sinh \left({\frac {x-C_{1}}{a}}\right)$

At x=14:

$\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right)$ .

b) The angle between the tangent to the catenary and the x-axis at a given point will be given by:

$\frac{dy}{dx}(x)=tan(\theta)$ ⇒ $\theta=tan^{-1} (\frac{dy}{dx}(x))$

Then, the angle between the line and the pole will be:

$\phi=\pi - \theta$ .

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4 years ago

Problem 12.78 Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi and

Gennadij [26K]

Answer:

M_e = 6.04 * 10^24 kg

Explanation:

Given:

- The time period of moon orbit T = 27.32 days

- Distance from center of earth to moon r = 238,910 miles

- The gravitational constant G = 6.67408 * 10^-11

Find:

- Determine the mass of the earth

Solution:

- The mass of earth as a function of time period T and radius of orbit r is related by an expression as follows:

M_e = (2*pi / T)^2 * (r^3 / G)

M_e = (2*pi / 27.32*24*60*60)^2 * (238910*1609 / 6.67408 *10^-11)

M_e = 6.04 * 10^24 kg

8 0

4 years ago