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avanturin [10]
2 years ago
9

If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

ng at constant speed across the floor with a constant, horizontal force of 70 N. The puppy has a weight of 110 N. What is the coefficient of static friction between the puppy and the floor
Physics
1 answer:
netineya [11]2 years ago
5 0

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

F_s = \mu_s*N

Where F_s is the static friction force, \mu_s is the coefficient of static friction and N is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, N = 110\text{ N}, to make the puppy moving we need to use a force of 80 N, therefore, F_s = 80 \text{ N}, so we can solve for the coefficient as shown below:

80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\

The coefficient of static friction between the puppy and the floor is 0.7273.

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2 years ago
Tammy leaves the office, drives 26 km due
fredd [130]

Answer:

72.98 km

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2 years ago
Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was s
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Answer

given,

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moving with speed = 11 m/s

time taken to stop = 0.14 s

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distance between Lisa ford and Kara's car = 30 m

a) change in momentum of Kara's car

  Δ P = m Δ v                  

  \Delta P = m (v_f-v_i)

  \Delta P = 1300 (0 - 11)

  Δ P = - 1.43 x 10⁴ kg.m/s

b) impulse is equal to change in momentum of the car

    I = - 1.43 x 10⁴ kg.m/s

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2 years ago
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Answer:

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this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

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let's replace

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2 years ago
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Answer:

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As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

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<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
2 years ago
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