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3 years ago

How long does it take an airplane to fly 5000 m if it maintains a speed of 240 meters per second?

Physics

1 answer:

kolbaska11 [484]3 years ago

8 0

Answer:

20 5/6 sec

Explanation:

To find the solution we divide 5000 by 240

However, when you see a problem, always try to simplify

5000/240=500/24=250/12=125/6

Now the division is much easier

20 5/6 sec

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Bob rides his bike with a constant speed of 10 miles per hour. How long will he take to travel a distance of 15 miles?

ICE Princess25 [194]

${\underline{\pink{\textsf{\textbf{ Answer : }}}}}$

➡ 150hrs.

${\underline{\pink{\textsf{\textbf{Explanation : }}}}}$

➡ Time = distance × speed

➡ Time = 15*10

➡ Time = 150hrs ans.

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2 years ago

Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1

harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum $p_{i}$ (before the elastic collision) must be equal to the final momentum $p_{f}$ (after the elastic collision):

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{1}V_{i} + m_{2}U_{i}$

$p_{f}=m_{1}V_{f} + m_{2}U_{f}$

Where:

$m_{1}=200 kg +100 kg=300 kg$ is the combined mass of Tubby and Libby with the car

$V_{i}=10 m/s$ is the velocity of Tubby and Libby with the car before the collision

$m_{2}=25 kg + 100 kg=125 kg$ is the combined mass of Flubby with its car

$U_{i}=0 m/s$ is the velocity of Flubby with the car before the collision

$V_{f}=4.12 m/s$ is the velocity of Tubby and Libby with the car after the collision

$U_{f}$ is the velocity of Flubby with the car after the collision

So, we have the following:

$m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f}$ (2)

Finding $U_{f}$ :

$U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}}$ (3)

$U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg}$ (4)

Finally:

$U_{f}=14.1 m/s$

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3 years ago

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Answer:

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