Momentum is a vector quantity that depends

How can Climate Change impact the formation of Hurricanes?

Orlov [11]

Climate change may affect tropical cyclones in a variety of ways: an intensification of rainfall and wind speed, a decrease in overall frequency, an increase in frequency of very intense storms and a poleward extension of where the cyclones reach maximum intensity are among the possible consequences

4 0

3 years ago

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Which sound waves in the electromagnetic spectrum are considered low energy waves

Mariulka [41]

Radio waves are the waves with the lowest energy in the electromagnetic spectrum. X-rays and gamma rays are the highest. Sound is not part of the electromagnetic spectrum.

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3 years ago

Which would be the best example to demonstrate the principle of conservation of energy? A) a stone is crushed B) light bends aro

Daniel [21]

The answer is a pendulum swings back an forth.

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3 years ago

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A marble at the front of a truck bed traveling at 20 m/s relative to the highway rolls toward the back of the truck bed with a s

irga5000 [103]

14 m/s in the direction of the truck

5 0

3 years ago

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$

7 0

3 years ago