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chubhunter [2.5K]
3 years ago
10

An athlete runs at a constant velocity of 5.2 m/s. What is the velocity of the athlete relative to the ground?

Physics
1 answer:
dimulka [17.4K]3 years ago
5 0

The relative velocity of the athlete relative to the ground is 5.2 m/s

The given parameters;

constant velocity of the athlete, V = 5.2 m/s

let the velocity of the ground = Vg = 0

The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.

The athlete is the moving object in this question while the ground is stationary.

The relative velocity of the athlete relative to the ground is calculated as follows;

V/V_g = V - V_g  = 5.2 - 0 = 5.2  \ m/s

Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s

Learn more here: brainly.com/question/24430414

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What happens to the appearance of an object as it gets hotter?
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Answer:

Explanation:

When an object is heated then it becomes brighter and bluish compared to the initial condition. This happens because when an object is given heat then the electron in the ground state gets excited and reaches some higher state. After reaching a higher state electron make the transition to lower state and simultaneously exhibit the color which is visible with naked eyes.

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The more light bulbs you add to a series circuit, the _______ the lights will become.
Virty [35]

Answer:

brighter

Explanation:

the more light bulbs you add to a series of circuits, the brighter the room will be.

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What is Planck’s law?
Lyrx [107]

Answer:

The wavelength of the emitted radiation is inversely proportional to its frequency, or λ = c/ν. The value of Planck's constant is defined as 6.62607015 × 10−34 joule∙second.

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Hope this helps!

3 0
3 years ago
1:
LenKa [72]

The answer of question 1 is:

C. The gravitational force becomes 16 times less.

The answer of question 2 is:

A. The gravitational force is 12 times larger.

The answer of question 3 is :

A. The new gravitational force is 9 times stronger than the old one.

The answer of question 4 is:

D. 8 times more.

6 0
3 years ago
The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe
shtirl [24]

Answer:

The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L

Explanation:

Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

5 0
3 years ago
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