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AlexFokin [52]
1 year ago
7

don is 10 kilometers ahead of jane, and both are traveling forward. how long will it take jane to catch up with don if jane is t

raveling at 10 k.p.h and don at 6 k.p.h. ?
Physics
1 answer:
swat321 year ago
8 0

The time it takes Jane to catch up with Don if Jane is traveling at 10 k.p.h and don at 6 k.p.h is: 2.5 h

The formula and procedure we will use to solve this problem is:

Tr = x/(v2-v1)

Where:

  • x = separation distance
  • Tr = reach time
  • v2 = velocity of the faster mobile
  • v1 = velocity of the slowest mobile

Information about the problem:

  • x = 10 km
  • v2 = 10 km/h
  • v1 = 6 km/h
  • Tr=?

Applying the reach time formula we get:

Tr = x/(v2-v1)

Tr = 10 km/(10 km/h- 6 km/h)

Tr = 10 km/(4 km/h)

Tr = 2.5 h

<h3>What is velocity?</h3>

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

Learn more about velocity at: brainly.com/question/80295?source=archive

#SPJ4

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Answer:

0.25 N

Explanation:

The density of the ball (\rho_b) = 0.0839 g/cm³ = 83.9 kg/m³

The density of water (\rho_w) = 1000 kg/m³

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radius of ball = 0.0377/2 = 0.01885 m

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Let us assume the acceleration due to gravity (g) = 9.8 m/s²

Hence:

The force is required to hold it completely submerged under water (F) is:

F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\

F = 0.25 N

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