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arlik [135]
3 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]3 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]3 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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A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho
d1i1m1o1n [39]

Answer:

Part a)

KE_r = 8 J

Part b)

v = 3.64 m/s

Part c)

KE_f = 12.7 J

Part d)

v = 2.9 m/s

Explanation:

As we know that moment of inertia of hollow sphere is given as

I = \frac{2}{3}mR^2

here we know that

I = 0.0484 kg m^2

R = 0.200 m

now we have

0.0484 = \frac{2}{3}m(0.200)^2

m = 1.815 kg

now we know that total Kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2

20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2

20 = 1.5125 v^2

v = 3.64 m/s

Part a)

Now initial rotational kinetic energy is given as

KE_r = \frac{1}{2}I(\frac{v}{R})^2

KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2

KE_r = 8 J

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

mgh = (KE_i) - KE_f

1.815(9.8)(0.900sin27.1) = 20- KE_f

7.30 = 20 - KE_f

KE_f = 12.7 J

Part d)

Now we know that

\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7

\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7

1.5125 v^2 = 12.7

v = 2.9 m/s

8 0
3 years ago
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Arada [10]

Answer:

15 watt

Explanation:

Power is the rate at which work is done.

This means you divide the work done with the amount of time used to perform the work.

The formula for Power is : P = W/t  where;

W= work done in J = 45

t= time in seconds = 3 sec

P= 45/ 3 = 15 watt

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Which is the force of repulsion between two positively-charged particles?
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<span> answer>>>>electric force <<<<by the way i don't like physics but i answer this for you ^-^</span>


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3 years ago
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