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max2010maxim [7]
3 years ago
9

A turtle, moving at a constant velocity of 0.9 m/s due East, is in a race with a rabbit, who runs at a moderate speed of 9 m/s.

When the turtle is 45 m from the finish line, the rabbit begins taunting the turtle by running from the turtle to the finish line (without crossing it) and back to the turtle. The rabbit continues going back and forth between the turtle and the finish line until the turtle crosses the finish line. About how many meters does the rabbit travel as the turtle travels that last 45 m? Assume the rabbit always runs at 9 m/s and doesn’t lose any time changing direction.
Physics
1 answer:
Nina [5.8K]3 years ago
7 0

Answer:

d = 450m

Explanation:

we know that:

V = d/t

where V is the velocity, d the distance and t the time.

so:

the turtle run at 0.9m/s a distance of 45m:

replacing v by 0.9 m/s and d by 45 m, we get:

V = d/t

0.9 = 45/t

solving for t:

t = 45/09 = 50 s

Now, using the same equation, the time that the turtle takes to travel 45 meters and the velocity of the rabbit, we get:

V = d/t

(9) = d/(50)

Solving for d:

d = 9*50 = 450m

it means that the rabbit travel 450m at the same time that the turtle travel 45m.

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kodGreya [7K]

Answer:

T = 15.03°C

Explanation:

given data:

copper specific heat = Sc = 0.385 J/g °C

iron specific iron = Si = 0.450 J/g °C

specific heat of ethanol = Se = 2.46 J/g °C

net heat loss is equal to zero

(m*S*\Delta T)_{copper} +(m*S*\Delta T)_ {iron} +(m*S*\Delta T)_ {ethanol} = 0

150*0.385 *( T - (-50)) + 200*0.450*(T - 120) + 300*2.46 * (T -20) = 0

57.75( T - (-50)) + 0.90(T - 120) +738(T -20) = 0

57.75T + 2887.5 + 0.90T - 108 + 738T - 14760 = 0

57.75T + 0.90T+738T = - 2887.5 + 108+14760

796.65T= 11980.5

T = 15.03°C

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3 years ago
if quasars often resemble little blue stars, what was it about them that so surprised astronomers when they were dis­covered?
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This condition, including other related evidence gotten in recent years concerning our galaxy, has shown that quasars are probably the central nuclei of very distant, very active galaxies.

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6 0
3 years ago
car travels 80 meters due north in 12 seconds then the car turns around and travels 30 Mi do South in 4 seconds calculate the av
Zinaida [17]

Answer:

1) 3.1 m/s

2) 7 m/s

Explanation:

Distance due north = 80 m

Distance due south = 30 m

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