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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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Discuss how the hardness or softness of the landing surface is related to the time required to stop the egg
monitta

Answer

To understand this concept it is necessary to understand Newton's Second Law

According to Newtons Second law applied force is equal to rate of change of momentum of a body.

Mathematically,

F=\frac{dp}{dt}

here, \frac{dp}{dt} is rate of change of momentum with respect to time

It means If two eggs fall from same height,one on softer surface and other on hard surface that time the momentum of both eggs will remain equal at both the surfaces. But, impact time will be different. On hard surface egg will stop almost instantly so impact time will be small and hence the force on egg will be large therefore the egg will breakup.

On the other hand on the soft surface like a cotton, egg will not stop instantly but it will slow down for a few seconds and then stop due to which the time of impact will increase. Therefore the force on egg will be less and it won't break up.


3 0
2 years ago
Think about it: suppose a meteorite collided head-on with mars and becomes buried under mars's surface. what would be the elasti
Ket [755]

Answer:

Perfectly inelastic collision

Explanation:

There are two types of collision.

1. Elastic collision : When the momentum of the system and the kinetic energy of the system is conserved, the collision is said to be elastic. For example, the collision of two atoms or molecules are considered to be elastic collision.

2. Inelastic collision: When the momentum  the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

For a perfectly elastic collision, the two bodies stick together after collision.

Here, the meteorite collide with the Mars and buried inside it, the collision is said to be perfectly inelastic. here the kinetic energy of a body lost completely during the collision.

4 0
2 years ago
Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal,
VARVARA [1.3K]

Answer: yes.

Explanation: The light that will be incidented on that metal is visible light.

It depends on 3 factors:

1. The temperature

2. The specific heat capacity of the metal

3. The thermal conductivity of the metal.

The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer. 

6 0
2 years ago
Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener
Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 \frac{m^{3} }{kg}

% Error =  20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 \frac{m^{3} }{kg}

% Error =  - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 \frac{J}{kg k}

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ \frac{V}{m} = \frac{R T}{P}

⇒ Here \frac{V}{m} = Specific volume = v

⇒ v =  \frac{R T}{P}

Put all the values in the above formula we get

⇒ v = \frac{323}{10^{6} } ×81.49

⇒ v = 0.02632 \frac{m^{3} }{kg}

This is the specific volume by ideal gas equation.

Actual value = 0.021796 \frac{m^{3} }{kg}

Error =  0.02632 - 0.021796 =   0.004524 \frac{m^{3} }{kg}

% Error =  \frac{0.004524}{0.021796} × 100

% Error =  20.75 %

(b). From the generalized compressibility chart the value of specific volume

 \frac{V}{m} = v = 0.0142 \frac{m^{3} }{kg}

The actual value = 0.021796 \frac{m^{3} }{kg}

Error = 0.0142 - 0.021796 =  \frac{m^{3} }{kg}

% Error = \frac{- 0.0076}{0.021796} × 100

% Error =  - 34.85 %

3 0
2 years ago
The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac
Leni [432]

Answer:

Maximum height the atmosphere pressure can support the

water=10.336 m

Explanation:

We know that ,

Pressure = h\cdot\rho\cdot g

Case 1 - Mercury in the tube

Density\ of\ mercury =\rho_1\\and\ height\ attained\ for\ mercury\ column = h_1

Case 2 - Water in the tube

Density\ of\ water =\rho_2\\and\ height\ attained\ for\ water\ column = h_2

Since atmospheric pressure is same

.P=h_1\cdot\rho_1\cdot g = h_2\cdot\rho_2\cdot g

or,  h_2=\frac{h_1\rho_1}{\rho_2}

Given\ h_1= 0.76\  m,\rho_1=13.6\cdot\rho_2

∴ h_2=0.76\cdot13.6=10.336\ m

Hence height of the water column =10.336 m

6 0
2 years ago
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