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charle [14.2K]
3 years ago
15

A student is bicycling north along Main Street to school. Another student is timing the bicycling student in order to determine

the average velocity. The bicycling student is 450 m away when the timer starts, and is 200 m away when the timer stops. If the student traveled the distance in 60 seconds, what is the students average velocity?
Physics
1 answer:
MatroZZZ [7]3 years ago
8 0

The average velocity is -4.17 m/s

Explanation:

The average velocity of a body is given by:

v=\frac{d}{t}

where

d is the displacement of the body

t is the time elapsed

For the student in this problem, we have:

Initial position: x_i = 450 m

Final position: x_f = 200 m

So the displacement is

d=x_f -x_i = 200 - 450 = -250 m

The time elapsed is

t = 60 s

Therefore, the average velocity is

v=\frac{-250}{60}=-4.17 m/s

Where the negative sign means the student is moving towards the origin.

Learn more about average speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

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natka813 [3]

Answer:

Explanation:

V₀ = 0 m/s

V = 41 m/s

S = 480 m

___________

a - ?

Distance traveled:

S =  (V² - V₀²) / (2*a)

Acceleration :

a = (V² - V₀²) / (2*S)

a = (41² - 0²) / (2*480) = 1.75 m/s²

8 0
1 year ago
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Ganezh [65]

Note that this is a position vs time graph.

From A to B, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

From B to C, the graph is a straight line with 0 slope. This indicates a constant position, i.e. the object remains stationary.

From C to D, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

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3 years ago
________ is a force acting through distance.
Andreas93 [3]
High school???
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What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the
Agata [3.3K]
The charge of the copper nucleus is 29 times the charge of one proton:
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the charge of the electron is
e=-1.6 \cdot 10^{-19}C
and their separation is
r=1.0 \cdot 10^{-12} m

The magnitude of the electrostatic force between them is given by:
F=k_e  \frac{Qe}{r^2}
where k_e is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N
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Answer:

8 times

Explanation:

Given that You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes.

That means the heart will pump 10 quarts in 2 minutes.

That is half of your blood volume per minute.

If during exercise it can pump 40 quarts per minute, that is, 80 quarts in 2 minutes.

To know how many times does all of your blood complete the cycle around your body during exercise, you must divide 80 quarts by 10 quarts. That is,

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