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Alina [70]
3 years ago
14

A heavy sled and a light sled, both moving horizontally with the same speed, suddenly slide onto a rough patch of snow and event

ually come to a stop. The coefficient of kinetic friction between the sleds and the rough snow is the same for both of them.
Which of the following statements about these sleds are correct?

A) Both sleds will slide the same distance on the rough snow before stopping.

B) The heavy sled will slide farther on the rough snow than the light sled.

C) The light sled will slide farther on the rough snow than the heavy sled.

D) The friction from the snow will do the same amount of work on both sleds.
Physics
2 answers:
Alex777 [14]3 years ago
5 0

Answer:

A) Both sleds will slide the same distance on the rough snow before stopping.

D) The friction from the snow will do more negative work on the heavy sled than on the light sled.

Explanation:The coefficient of kinetic friction is the ratio of the force of friction of a given object or material to the normal force acting on it,it depends mainly on the nature of the surface in consideration. The higher the roughness of a surface is, the higher the coefficient of kinetic friction. For an object at rest on a plane surface with no other force acting on it,the normal force will be considered as the force of gravity.

maria [59]3 years ago
4 0

Answer:

The correct answers is option A) "Both sleds will slide the same distance on the rough snow before stopping".

Explanation:

For objects moving horizontally at the same speed, an equal coefficient of kinetic friction would mean that both objects will move the same distance before stopping, independently of their mass. The coefficient of kinetic friction is obtain from dividing the force needed to pull an object between the force that holds or prevents the object from moving. The coefficient of kinetic friction is a dimensionless scalar, which means that is independent of the size of the object because it depends only in the forces that act upon it.

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Answer:

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Explanation:

given data

oven = 350◦F

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solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

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