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Alina [70]
3 years ago
9

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

t 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?
A. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase.B. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.C. The magnitude of the normal force is equal to the sum of the magnitude of the force of the pull and the magnitude of the suitcase's weight.D. The magnitude of the normal force is greater than the magnitude of the weight of the suitcase

Physics
2 answers:
Stolb23 [73]3 years ago
8 0

Answer:

The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.

Explanation:

aleksandr82 [10.1K]3 years ago
4 0

Answer:B

Explanation:

Given

man is unable to lift the suitcase thus

Magnitude of Normal reaction will  weight minus magnitude of force pull

From FBD

N+F=W_s

N=W_s-F

Where F=Pull Force

N=Normal reaction

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Answer:

0.00000019631 meters

Explanation:

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3 years ago
Name the physical quantity measured by the velocity-time graph?
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The voltage across the diode indicates the energy given to charge carriers (electrons and holes, but more about that later in th
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Answer:

the charge carriers have an energy 2.8 10⁻¹⁹ J

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3 years ago
A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta
larisa [96]

Answer:

Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

x = 26.16 m

Part d)

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Part e)

x = 83.23 m

Part f)

y = -75.6 m

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

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similarly we have

v_y = 19.9 sin39.9

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Part a)

Horizontal displacement in 1.03 s

x = v_x t

x = (15.3)(1.03)

x = 15.76 m

Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

y = 7.94 m

Part c)

Horizontal displacement in 1.71 s

x = v_x t

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x = 26.16 m

Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

y = 7.49 m

Part e)

Horizontal displacement in 5.44 s

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Part f)

Vertical direction we have

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y = (12.76)(5.44) - 4.9(5.44)^2

y = -75.6 m

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