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3 years ago

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gladu [14]3 years ago

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Your question did not get pasted correctly. Please take a photo of the question and add to the original post. I will correct my answer once that is done. Thank you.

almond37 [142]3 years ago

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Explanation: dont put fake answers on people's pages.

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Which of the following is a sign of a physical Change ?

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C.) i believe because the change in shape is changing the physical appearance

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4 years ago

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A nerve signal travels 150 meters per second. Determine the number of kilometers that the nerve signal will travel in the same t

joja [24]

Explanation:

It is given that,

A nerve signal travels 150 meters per second. It is the speed of the nerve signal. We need to convert the number of kilometers that the nerve signal will travel in the same time.

We know that,

1 kilometer = 1000 meter

1 hour = 3600 seconds

$150\ \dfrac{m}{s}=150\times \dfrac{1/1000\ km}{1/3600\ h}$

$150\ m/s=540\ km/h$

So, the nerve signal will travel at the rate of 540 km/h. Hence, this is the required solution.

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3 years ago

Which of the following is not an example of projectile motion?

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4 years ago

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An airplane traveling from Atlanta to New York travels 1,395 km in 2.5 h. What is the plane’s average speed?

Roman55 [17]

D. 558 km/h because if you divide 1,395 by 2.5 you’ll get the answer and I know this is correct because the keyword for me to divide is average.

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3 years ago

A 0.5 kg block slides down a frictionless semicircular track from a height, h, and collides with a second 1.5 kg block at the bo

Ede4ka [16]

Answer:

$h' = 0.062\cdot h$

Explanation:

The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:

$U_{g} = K$

$m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot h}$

$v \approx 4.429\cdot \sqrt{h}$

Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:

$(0.5\,kg) \cdot (4.429\cdot \sqrt{h}) = (2\,kg)\cdot v$

$v = 1.107\cdot \sqrt{h}$

Lastly, the height reached by the two blocks is:

$K = U_{g}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot h'$

$h' = \frac{v^{2}}{2\cdot g}$

$h' = \frac{(1.107\cdot \sqrt{h})^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h' = 0.062\cdot h$

3 0

3 years ago