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klemol [59]
1 year 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]1 year 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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Debora [2.8K]

Answer:

The gravitational potential energy equals the work needed to lift the object.

Explanation:

here we know that

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W = \vec F . \vec d

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force due to gravity is given as

\vec F_g = mg

now here if we plug in the value of distance and force in the formula of work done then we will have

W = (mg)(h)

so here we got

W = PE_g

so we can concluded that

The gravitational potential energy equals the work needed to lift the object.

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1 year ago
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An object is 6.0 cm in front of a converging lens with a focal length of 10 cm.Use ray tracing to determine the location of the
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As we know that

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

here we know that

d_0 = 6 cm

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now from above equation we have

\frac{1}{d_i} + \frac{1}{6} = \frac{1}{10}

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Ray diagram is attached below here

8 0
2 years ago
How far could a rabbit run if it ran 36km/h for 5.0min?
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A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi
RSB [31]

Answer:

r = 4.44 m

Explanation:

 

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

         B = ρ g V

Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

         σ = W / A

         W = σ A

The area of ​​a sphere is

           A = 4π r²

       W = W₁ + σ 4π r²

The volume of a sphere is

           V = 4/3 π r³

Let's replace

     ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

     P V = n RT

    P = ρ RT

    ρ = P / RT

 

    P / RT g 4/3 π r³ - σ 4 π r² = W₁

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Let's replace the values

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     r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

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3 0
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Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp
UkoKoshka [18]

Answer:

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Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

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