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Veseljchak [2.6K]

3 years ago

9

In the picture at left. The launch height is 10m ball B mass 5kg has a launch velocity of 12 m/s and ball A mass 8 kg has a laun

ch velocity twice that determine the time in flight for both the range for each and explain why the difference is masses is not considered

1 answer:

Hitman42 [59]3 years ago

8 0

Answer:

24

Explanation:

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Drag the tiles to the correct boxes to complete the pairs.

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Air pollution---> smog

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3 years ago

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egoroff_w [7]

B is the correct answer since acid corrode on metals such as carbon, steel, zinc and such.

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3 years ago

If the car speeds up at a steady 1.6 m/s2 , how long after starting is the magnitude of its centripetal acceleration equal to th

Rufina [12.5K]

Based on internet sources, <span>the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a).
</span>
<span>Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. </span>
<span>Acceleration = 1.6 cramwells/s^2 </span>
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I hope this helps.</span>

8 0

3 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

a)

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

During which phase of the moon do neap tides occur?

Fynjy0 [20]

Answer:

First Quarter and Third Quarter.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.

Since gravity variates with the distance:

$F = G\frac{m1\cdot m2}{r^{2}}$ (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).

However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.

Therefore, that happens when the Moon is at First Quarter and Third Quarter.

4 0

3 years ago

Read 2 more answers