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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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Number of waves that pass a given point in one second

Studentka2010 [4]

<em>number of waves that pass a given point in one second is called <u>frequency..</u></em>

5 0

3 years ago

Present day glaciers are found primarily in _______________.

Mazyrski [523]

<h3><u>Answers;</u></h3>

Antarctica and Greenland

Present day glaciers are found primarily in <em><u>Antarctica and Greenland</u></em>.

<h3><u>Explanation;</u></h3>

<em><u>The two major ice sheets that exists today are found primarily in Antarctica and Greenland. Ice sheets are large masses of glacial ice that are also known as continental glaciers.</u></em>
Most ice in Antarctica and Greenland spill out into the ocean from a few spots. The Antarctica and Greenland ice sheets combined comprise more than 99 percent of freshwater ice found on Earth.

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3 years ago

Read 2 more answers

Suppose a car approaches a hill and has an initial speed of

kvv77 [185]

Answer:

a) 1.73*10^5 J

b) 3645 N

Explanation:

106 km/h = 106 * 1000/3600 = 29.4 m/s

If KE = PE, then

mgh = 1/2mv²

gh = 1/2v²

h = v²/2g

h = 29.4² / 2 * 9.81

h = 864.36 / 19.62

h = 44.06 m

Loss of energy = mgΔh

E = 780 * 9.81 * (44.06 - 21.5)

E = 7651.8 * 22.56

E = 172624.6 J

Thus, the amount if energy lost is 1.73*10^5 J

Work done = Force * distance

Force = work done / distance

Force = 172624.6 / (21.5/sin27°)

Force = 172624.6 / 47.36

Force = 3645 N

5 0

2 years ago

What are the two factors that determine an objects gravitational potential energy

Savatey [412]

<span>-Mass (kg)
-Velocity (m/s)</span>

8 0

2 years ago

What is the strength of the electric field 3.0 cm from a small glass bead that has been charged to + 8.0 nc ?

Debora [2.8K]

Thinking the small glass bead as a single point charge, the electric field generated by it is given by
$E(r) = k_e \frac{Q}{r^2}$
where
$k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant
$Q=8.0 nC=8.0 \cdot 10^{-9} C$ is the charge of the bead
$r=3.0 cm=0.03 m$ is the distance at which we calculate the field.

Using these data, we find:
$E=(8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{(8.0 \cdot 10^{-9} C}{0.03 m)^2}=8.0 \cdot 10^4 N/C$

6 0

2 years ago