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zubka84 [21]
3 years ago
8

A speedy tortoise can run with a speed of

bsmiddle" class="latex-formula"> and a hare can run with a speed v_H. The hare waits to rest for a time Δt₀ after the race begins and the tortoise starts and then runs as fast as he can, but the tortoise still wins by a shell (length of shell = s). The length of the race track is a distance, d. The race is considered over when the tortoise crosses the finish line.
Using the symbols of the problem, write an equation for the amount of time that the hare ran, \Delta t_H.
Physics
1 answer:
Vinil7 [7]3 years ago
8 0

Answer:

t_{H}  = t_{T}  - 60

Explanation:

Thinking process:

Let:

distance = speed × time

for the hare:

d_{H}  = s_{H} t_{H}

for the tortoise:

d_{T}  = s_{T} t_{T}

Let the length of the shell be s. Then the two are related. So:

d_{H}  = d_{T}  - s

Let's say the hare ran at x times the speed of the tortoise:

s_{H}  = xs_{T}

So, the equation still holds : t_{H}  = t_{T}  - 60

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A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi
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Answer:

Explanation:

a ) Time period  T = 2 s

Angular velocity ω = 2π / T

=  2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

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Final moment of inertia

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7 0
3 years ago
DUE BY MIDNIGHT
olga2289 [7]

Answer:

Option D. 1000 J.

Explanation:

From the question given above, the following data were obtained:

Force (F) applied = 200 N

Distance (s) = 5 m

Time (t) = 10 s

Workdone (Wd) =?

Workdone (Wd) is simply defined as the product of force (F) and distance (s) moved in the direction of the force. Mathematically, it is expressed as:

Wd = F × s

With the above formula, we can calculate the Workdone as illustrated below:

Force (F) applied = 200 N

Distance (s) = 5 m

Workdone (Wd) =?

Wd = F × s

Wd = 200 × 5

Wd = 1000 J

Thus, the Workdone is 1000 J

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