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3 years ago

7

What is a wave coming into a barrier

1 answer:

3241004551 [841]3 years ago

6 0

A wave looses its power as it comes to shore because it gets less deeper every second it gets closer to shore

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PLEASE HELP AND HURRY!!!!!!!!!

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I believe the answer would be A. point x

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3 years ago

The Netherlands controlled ______ for centuries <br><br> A. Vietnam<br> B. Malaysia<br> C. Indonesia

gogolik [260]

The Netherlands controlled Indonesia for centuries!

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3 years ago

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If someone drove for 20 minutes going 30 m/s, how far did they drive?

vladimir2022 [97]

D- 36,000m is the answer

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3 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 100 m in 25 s?

larisa86 [58]

As per the question, the distance travelled by bobsled [s] = 100 m

The time taken by the bobsled to travel that distance [t] = 25 s

We are asked to calculate the speed of the bobsled.

The speed of the bobsled is calculated as -

$speed\ =\ \frac{distance}{time}$

$v\ =\ \frac{s}{t}$

$=\frac{100\ m}{25\ s}$

$=\ 4\ m/s$

Hence, the correct answer to the question is A. 4 m/s.

8 0

3 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:

$U_{f}=14.1 m/s$

8 0

3 years ago