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2 years ago

25 points please help PLEASE

Physics

1 answer:

Nadusha1986 [10]2 years ago

3 0

Answer:

você tem um camping de em Rio do Rio de Janeiro ou em lojas em SP e SP e RJ e região metropolitana de Porto Velho de Porto Velho Rio

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12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and

Nimfa-mama [501]

Answer:

I think 9.5

Explanation:

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3 years ago

A brown bear runs at a speed of 9.0\,\dfrac{\text m}{\text s}9.0 s m 9, point, 0, start fraction, start text, m, end text, div

KengaRu [80]

Explanation:

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3 years ago

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This back-and-forth movement of electrons is called . In contrast, the movement of electrons in one direction in a battery circu

stich3 [128]

The back-and-forth movement of electrons is called alternating current. Electrons go back and forth, the direction of their path alternates from one direction to another.

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3 years ago

What is the difference between biopsychology and neuropsychology?

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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:

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

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3 years ago