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pogonyaev
3 years ago
11

3. A cart at rest at the top of a roller coaster has 1000 J of

Physics
2 answers:
vovikov84 [41]3 years ago
5 0

Answer:

600J

Explanation:

Assuming no energy is lost and perfect transfers.

On the top of the first hill, E_{k}+E_{p}=1000 - however there is no Kinetic Energy so - E_k = 0

On the second hill, the total amount of energy is still the same (1000) - this time we have a smaller amount of potential energy. We can rearrange to get E_k=1000-400

wariber [46]3 years ago
4 0

Answer:

j

Explanation:

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A race car drives one lap around a race track that is 500 meters in length.
pychu [463]

Explanation:

Check out the picture I drew for a minute before reading this...

B. Distance [the red line] is a scalar quantity reflecting how far an object has traveled. Displacement [the green line] is a vector quantity reflecting how far an object has moved from a point. The key difference is that distance can be any sort of path while displacement is always a vector (or a straight line) between a starting point and a finishing point. Sometimes distance and displacement are equal to one another. Sometimes you have a distance traveled, but zero displacement overall; which is what's going on in your question.

A. The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.

3 0
3 years ago
SHOW WORK
Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-ha
vovikov84 [41]
If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.

The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.
6 0
3 years ago
Read 2 more answers
What is the velocity of a 0.8kg ball that has a momentum of 3 kg*m/s?
Marysya12 [62]

The velocity is 3.75

5 0
3 years ago
He electron cloud is the predicted location for
DanielleElmas [232]

Answer:

A) Electrons only

Explanation:

In quantum mechanics, "electron cloud" refers to the region of space where electrons can be found in an atom. In fact, in quantum mechanics, it is not possible to predict the exact position of the electrons, so there is a region where electrons have a certain probability to be found.

7 0
3 years ago
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