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yKpoI14uk [10]
3 years ago
9

In a “minute to win it” game, cards are placed between cups to stack them. The contestant then pulls the card out in hopes that

they will end with a set of stacked cups. Why does this work?
There is no friction between the card and the cup.


Inertia will keep the top cup from moving very far. Gravity will pull one cup on top of another.


The opening of the cup is so much bigger than the bottom, even though the cup moves they will both be stacked together.


Both cups are moved the same amount by the friction between them and the card.
Physics
2 answers:
Luden [163]3 years ago
4 0

Answer:

There is no friction between the card and the cup.

Explanation:

lubasha [3.4K]3 years ago
3 0

Answer:

The answer option to your question is A) - There is no friction between the card and the cup.

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Determine which heat transfers below are due to the process of conduction. I) You walk barefoot on the hot street and it burns y
Taya2010 [7]

Answer:

I) You walk barefoot on the hot street and it burns your toes.

II) When you get into a car with hot black leather in the middle of the summer and your skin starts to get burned.

Explanation:

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Here due to this direct contact the energy is transferred via a given solid or liquid medium

In this type of heat transfer medium particles will remain in its own position only the energy is transferred.

So here we can say the correct answer will be

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3 0
3 years ago
A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte
vaieri [72.5K]

Answer:

Part a)

H = 44.1 m

Part b)

y = 13.48 m

Part c)

d = 8.86 m

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

t = \frac{v sin\theta}{g}

now we have

3 = \frac{vsin\theta}{9.8}

v sin\theta = 29.4 m/s

Now maximum height above ground is given as

H = \frac{v^2sin^2\theta}{2g}

H = \frac{29.4^2}{2(9.8)}

H = 44.1 m

Part b)

Height of the fence is given as

y = (vsin\theta) t - \frac{1}{2}gt^2

y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)

y = 13.48 m

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

x = v_x t

97.5 = v_x (5.5)

v_x = 17.72 m/s

now total time of flight is given as

T = 3 + 3 = 6 s

so range is given as

R = v_x T

R = (17.72)(6)

R = 106.4 m

so the distance from the fence is given as

d = 106.4 - 97.5

d = 8.86 m

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