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Goryan [66]
3 years ago
8

Two balls move down a incline. One spins, while the other slides. Which one is going faster at the bottom?

Physics
1 answer:
natta225 [31]3 years ago
7 0

Answer: the one spinnin

Explanation:

You might be interested in
What do we have seasons on earth
Goryan [66]

Because the Earth's axis is not "straight up and down" as we move
around the sun. 

So when we're on one side of the sun, the top pole leans slightly toward
the sun.  During that time the sun shines more directly on the top half
of the Earth, and less directly on the bottom half.  The people on the
top half see the sun higher in the sky, and their weather is warmer,
while the people on the bottom half see the sun lower in the sky, and
their weather is cooler.

Then, when we're on the other side of the sun, the top pole leans slightly
away from the sun.  During that time the sun shines more directly on the
bottom half of the Earth, and less directly on the top half.  The people on
the bottom half see the sun higher in the sky, and their weather is warmer,
while the people on the top half see the sun lower in the sky, and their
weather is cooler.

The Earth makes the complete trip around the sun in one year, so the
people on the Earth go through this cycle of higher/lower sun and
warmer/cooler weather every year.

8 0
3 years ago
A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d
QveST [7]

Answer:

  The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

  (target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

  (10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

  (target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

  (target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

  -94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

  13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

  23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

  (12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

  target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

  √(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0
3 years ago
Which statement best describes the isothermal process? A. the temperature remains constant B. the temperature increases at a con
Veronika [31]

Answer:

The answer would be A. - the temperature remains constant

Explanation:

An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0

5 0
3 years ago
Read 2 more answers
Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho
uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v   proportional to  \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

{\frac{M_{B}}{M_{A}}} = 4

{\frac{M_{B}}{4}} = M_{A}

7 0
3 years ago
A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.
Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is  1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

#SPJ1

6 0
2 years ago
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