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lesya692 [45]
3 years ago
15

you nose out another runner to win 100.000 m dash. if your total time for the race was 13.800 s and you aced out the other runne

r by 0.001 s, by how many meters did you win?
Physics
1 answer:
sammy [17]3 years ago
3 0

Your average speed was

(100 m) / (13.8 s) = 7.25 m/s .

If you finished 0.001s ahead of him, then at your average speed, that corresponds to

(7.25 m/s) x (0.001 s) = 0.00725 m

That's 7.25 millimeters ... about 0.28 of an inch !

NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.

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2 years ago
Describe what happens to the electric field lines when two objects with unlike charges are brought near each other.
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The answer is:

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4 0
3 years ago
Read 2 more answers
A sled is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy isJ. (Formula: PE =
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Data:

h = 2m
m = 45 Kg
PE = ? (Joule)

Adopting, gravity (g) ≈ <span>9,8 m/s² 
</span>
Formula: PE_{grav}  = mass * g * height

Solving:

PE_{grav} = mass * g * height
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PE_{grav} = 882J

Answer:
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8 0
2 years ago
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What influences the strength of an electric field?
Slav-nsk [51]

Answer:

Explanation:

The charge alters that space, causing any other charged object that enters the space to be affected by this field. The strength of the electric field is dependent upon how charged the object creating the field is and upon the distance of separation from the charged object.

4 0
3 years ago
A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.
Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

d=30,000 pc =9.26\cdot 10^{20} m

The speed of the cosmic ray is

v=0.98 c

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c=3.0 \cdot 10^8 m/s is the speed of light. Substituting,

v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s

Converting into years,

T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}

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T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years

3 0
3 years ago
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