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arlik [135]
4 years ago
15

A football player runs from his own goal line to the opposing team's goal line returning to the fifty yard line all i 18.0s calc

ulate average speed and average velocity
Physics
2 answers:
ANTONII [103]4 years ago
3 0
<span>assuming the pitch is 100yards long, the player runs 100yards to the other goal then a further 50 yards back to the 50-yard line. So he/she runs 150yards in 18s
150/18 = 8.33yards per second average speed.
Initial velocity = 0, average velocity =8.33
Vav = (Vinitial+Vfinal)/2
Vav = 4.16m/s</span>
mylen [45]4 years ago
3 0

Answers:

Average speed = 8.33 yard/s

Average velocity = 2.78 yard/s

Explanation:

1) Definitions and equations:

i) Average speed = distance run / time elapsed

ii) Average velocity = displacement / time elapsed

iii) displacement = [final position - initial position]

2) Distance run:

i) Take a total length of 100 yards from goal to goal lines

ii) distance run = distance from goal to goal lines + back to yard 50 = 100 yards + 50 yards = 150 yards

3) Average speed = distance run / total time = 150 yards / 18.0s = 8.33 yard/s

4) Displacement:

final position - initial position = 50 yard - 0 yard = 50 yards

5) Average velocity = 50 yards / 18.0 s = 2.78 yards/s

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5 0
3 years ago
Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia
USPshnik [31]

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

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8 0
3 years ago
A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,
Marina CMI [18]

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. <em>The correct form is: A </em>+10\,\mu C<em> charge and a </em>-10\,\mu C<em> at a distance of 0.3 meters. </em>

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}} (1)

Where:

F - Electrostatic force, in newtons.

\kappa - Electrostatic constant, in newton-square meters per square coulomb.

|q_{A}|,|q_{B}| - Magnitudes of electric charges, in coulombs.

r - Distance between charges, in meters.

If we know that \kappa  = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, |q_{A}| = |q_{B}| = 10\times 10^{-6}\,C and r = 0.3\,m, then the magnitude of the electrostatic force is:

F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}

F = 9.987\,N

In consequence, correct answer is B.

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Explanation:

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It depends if they have the same lightbulb in them.

Explanation:

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