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attashe74 [19]
3 years ago
7

A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. (See Figure 1) However, he is unable t

o lift the suitcase from the floor. Which statement about the magnitude of the normal force n acting on the suitcase is true during the time that the man pulls upward on the suitcase?
Physics
1 answer:
cupoosta [38]3 years ago
3 0

Answer:

Normal force, N = W_s - F

Explanation:

Let w_s is the weight of suitcase. A man attempts to pick up his suitcase by pulling straight up on the handle. The weight of the suitcase in downward direction. The normal force is acting in upward direction. Let F is the force with which it is pulled straight up.

So, the normal force is given by :

N = W_s - F

N = mg - F

mg is the weight of the suitcase.

Hence, this is the required solution.

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Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
If you were told an atom was an ion, you would know the atom must have a ______?
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charge

Explanation:

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a shopper pushes a cart 40.0m south down one aisle and then turns 90.0 degrees and moves 15.0m. He then makes another 90.0 degre
valentinak56 [21]

Answer:

Explanation:

The displacement is the distnce of the shopper from the starting point.

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