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klemol [59]
2 years ago
12

In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move

a bit farther across the ice and other times players want the stones to stop a bit sooner. Suggest a way to increase the kinetic friction between the stone and the ice so that the stone stops more quickly. Next, suggest a way to decrease the kinetic friction between the stone and the ice so that the stone slides farther along the ice before coming to a stop
Physics
1 answer:
lara31 [8.8K]2 years ago
8 0

Answer:

To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.

To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.

Explanation:

In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.

On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.  

The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.

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The A-string (440 HzHz) on a piano is 38.9 cmcm long and is clamped tightly at both ends. If the string tension is 667N, what's
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Answer:

Mass, m = 2.2 kg                                

Explanation:

It is given that,

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Tension in the spring, T = 667 N

The frequency in the spring is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

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On rearranging, we get the value of m as follows :

m=\dfrac{T}{4Lf^2}

m=\dfrac{667}{4\times 0.389\times (440)^2}

m = 0.0022 kg

or

m = 2.2 grams

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2 years ago
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Answer:

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From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

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Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

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Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

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Answer:

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