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3 years ago

How does the numerical value of "e" change as the shape of the ellipse approaches a straight line?

1 answer:

4 0

At eccentricity = 0 we get a circle For 0 < eccentricity < 1 we get an ellipse for eccentricity = 1 we get a parabola for eccentricity > 1 we get a hyperbola for infinite eccentricity we get a lineSHOW FULL ANSWER

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How was the speed of light determined

goblinko [34]

Presently, the speed of light in a vacuum is defined to be exactly 299,792,458 m/s (approximately 186,282 miles per second). . An early experiment to measure the speed of light was conducted by Ole Romer, a Danish physicist, in 1676. Using a telescope, Ole observed the motions of Jupiter and one of its moons, Io

5 0

3 years ago

Read 2 more answers

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

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3 years ago

A diverging mirror is .................

lyudmila [28]

Is there a multiple choice?

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4 years ago

It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibilty can be described by Hooke

Varvara68 [4.7K]

<span>k = 1.7 x 10^5 kg/s^2 Player mass = 69 kg Hooke's law states F = kX where F = Force k = spring constant X = deflection So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration. F = kX F/X = k 115 kg* 9.8 m/s^2 / 0.65 cm = 115 kg* 9.8 m/s^2 / 0.0065 m = 1127 kg*m/s^2 / 0.0065 m = 173384.6154 kg/s^2 Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2 Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So X/0.39 cm = 115 kg/0.65 cm X = 44.85 kg/0.65 X = 69 kg The player masses 69 kg.</span>

4 0

4 years ago

Read 2 more answers

PLEASE HELP WILL GIVE BRAINLEST!!!

Aleks04 [339]

Answer:

huh what do u want tho its nothing bad

Explanation:

3 0

3 years ago