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2 years ago

At the 1 s mark, which objects have equal momentum?

Physics

1 answer:

Rudik [331]2 years ago

3 0

C. The bowling ball and the bicycle

p = mv
P of bike = 12x5 = 60
P of rock = 2x20 = 40
P of ball = 5 x 10 = 60

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For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity $e$ of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

$e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}$

Where:

$r_{a}$ is the apoapsis (the longest distance between the moon and its planet)

$r_{p}=0.27 r_{a}$ is the periapsis (the shortest distance between the moon and its planet)

Then:

$e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}$

$e=\frac{0.73 r_{a}}{1.27 r_{a}}$

$e=0.57$ This is the moon's orbital eccentricity

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3 years ago

The traditional view of mangrove forests as wastelands and unhealthy environments helped promote their degradation because _____

skad [1K]

Raditionally mangrove forests were viewed as wastelands and unhealthy environments. People therefore reasoned that their removal would improve the health ...

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2 years ago

A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50

kolezko [41]

Answer:

$Distance_{vaccum}=5.19cm$

Explanation:

The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under

Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

$T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}$

$T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}$

Thus in the same time the it would had covered a distance of

$Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]$

we have

$\mu _{ice}=1.309\\\\\mu _{quartz}=1.542$

Applying values we have

$Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]$

$Distance_{vaccum}=5.19cm$

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