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Oliga [24]
2 years ago
15

If the average velocity during the athlete's walk back

Physics
1 answer:
goblinko [34]2 years ago
4 0

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

T=\frac{d}{v}

Therefore

T=\frac{d}{1.50}

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

For more information on this visit

brainly.com/question/22271063

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2 years ago
A truck travels up a hill with a 7.5◦incline.The truck has a constant speed of 24 m/s.What is the horizontal component of thetru
riadik2000 [5.3K]

Answer:

The horizontal component of the truck's velocity is: 23.70 m/s

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The identities are:

Cosα= \frac{CA}{H}

Senα= \frac{CO}{H}

Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus

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Let Vx represent it.

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The vertical component of the truck's velocity is:

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