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Yakvenalex [24]
3 years ago
10

cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your

answer in seconds, minutes, and hours
Physics
1 answer:
sp2606 [1]3 years ago
4 0

We have that the time in seconds, minutes, and hours is

t=1.083*10^{-4}s

T_{min}=1.805*10^{-6}min

T_{hours}=3.0083*10^{-8}hours

From the Question we are told that

Velocity v=13.0m/s

Distance d=120 km

Generally the equation for the Time  is mathematically given as

t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s

Therefore

T_{min}=1.083*10^{-4}s/60

T_{min}=1.805*10^{-6}min

And

T_{hours}=T_{min}/60

T_{hours}=3.0083*10^{-8}hours

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

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When the moon is between the earth and the sun what moon phase will this be
Aleksandr [31]

Answer:  The new moon phase occurs when the Moon is directly between the Earth and Sun. A solar eclipse can only happen at new moon. A waxing crescent moon is when the Moon looks like crescent and the crescent increases ("waxes") in size from one day to the next. This phase is usually only seen in the west.

3 0
3 years ago
A ball bearing is projected vertically upwards from the ground with a velocity of 15ms. Calculate the time taken by the ball to
cupoosta [38]

Answer:

t = 3 [s]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}-g*t

where:

Vf = final velocity [m/s]

Vo = initial velocity = 15 [m/s]

g = gravity acceleration = 10 [m/s²]

t = time [s]

Now replacing we have:

0 = 15 -10*t\\10*t=15\\t= 1.5[s]

Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.

The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:

t = 1.5 + 1.5 = 3 [s]

7 0
3 years ago
You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
wlad13 [49]

Answer:

  • between locations that are 14 cm outboard of the chair edges
  • the weightless board is centered and end sitters are 25 cm from the ends

Explanation:

We can assume the .5 m-wide chair means that it is comfortable for each student to sit 0.25 m from the end of the board. If the board is centered on the chair, then each student is 1 m from the edge of the chair.

When Dan and Tahreen are seated on the board, their center of mass is ...

  (50 kg×2.5 m)/(50 kg +67 kt) = 1.068 m

to the right of the position where Dan is seated. Since this location is over the chair, the board is stable.

Komila can sit as much as x distance from the chair toward Dan, where ...

  67(1) +54(x) = 50(1.5)

  x = 8/54 ≈ 0.148 . . . . meters

Or, Komila can sit as much as x distance from the chair toward Tahreen, where ...

  67(1.5) = 54(x) +50(1)

  x = 50.5/54 ≈ 0.935 . . . . meters

<u>Scenario 1</u>

Assuming the (weightless) board is centered on the chair, Komila can sit anywhere between 14.8 cm left of the chair and 93.5 cm right of the chair and the board will remain stable. Sitting on the board centered on the chair is a suitable location. The two students sitting on the ends must become (and stay) seated at the same time. They both must be seated 0.25 m from the end of the board for the other dimensions to remain valid.

<u>Scenario 2</u>

Assuming the (weightless) board is located so its left end is 1.068 m from the chair, and Dan and Tahreen are seated 0.25 m from the ends of the board, Komila can sit anywhere within (117/54×.25 m) = 0.54 m of the chair and the board will remain stable. Again, sitting centered on the chair is a suitable location.

__

There does not appear to be any location where Komila can sit and have the board remain stable with only Dan or Tahreen seated on one end (assuming a width of 0.5 m for each sitter).

_____

<em>Comment on the question</em>

For the board to remain stable, the sum of moments about either edge of the chair must tend to rotate the board toward the chair. This sum will depend on the locations of the sitters relative to each edge of the chair, so there is significant freedom in choosing locations. To make the problem tractable, we have made some specific assumptions about where the board is and what the locations of the sitters might be. YMMV

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What unit of time on Earth is based on the revolution of the Earth around the Sun?
ollegr [7]

C.) Year

i hope this helped! please mark me as brainliest and 5 stars >:)

3 0
3 years ago
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Bob is coming toward you at a speed of 75 km/hr. You throw a baseball in his direction at 75 km/hr. What does he see the ball do
KengaRu [80]

Answer:

He sees the ball coming at him at 150 km/hr.

Explanation:

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Due to the same direction, both velocities will be added.

4 0
4 years ago
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