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zubka84 [21]
2 years ago
6

A man attempts to pick up his suitcase of weight w_s by pulling straight up on the handle. (Part A figure) However, he is unable

to 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?
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

Now assume that the man of weight w_m is tired and decides to sit on his suitcase. (Part B figure) Which statement about the magnitude of the normal force n acting on the suitcase is true during the time that the man is sitting on the suitcase?
Physics
1 answer:
alexira [117]2 years ago
8 0

Answer:

Part A. 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.

Part B. The magnitude of normal force acting on the suitcase is equal to the sum of the weight of the suitcase and the man.

Explanation:

Part A. This is because when the man pulls on the suit upwards, he exerts a force in the upward direction. This takes part of the force of weight of the suitcase and decreases the force the suitcase is exerting on the ground. Thus, the normal force (force exerted by suitcase on the ground) also decreases by the same force as the pull.

Part B. The statements for this part were not given in the question, but the answer reflects what is going to happen in that scenario. Since the man sits on the suitcase, the total weight acting on the ground through the suitcase is that of the suitcase plus the man. Since this force (acting on the ground) is normal force, the statement given in the answer is correct.

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A ladder placed up against a wall is sliding down. The distance between the top of the ladder and the foot of the wall is decrea
Kitty [74]

Answer:

distance changing at rate of 3.94 inches/sec

Explanation:

Given data

wall decreasing at a rate = 9 inches per second

ladder L = 152 inches

distance  h = 61 inches

to find out

how fast is the distance changing

solution

we know that

h² + b² = L²   ..................1

h² + b² = 152²

Apply here derivative w.r.t. time

2h dh/dt + 2b db/dt = 0

h dh/dt + b db/dt = 0

db/dt = - h/b × dh/dt     .............2

and

we know

h = 61

so h² + b² = L²

61² + b² = 152²

b² = 19383

so b = 139.223

and we know dh/dt = -9 inch/sec

so from equation 2

db/dt = -61/139.223  (-9)

so

db/dt = 3.94 inches/sec

distance changing at rate of 3.94 inches/sec

3 0
2 years ago
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
Alja [10]

Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = 12.0v+4.00v^2

Equating both, we have

    784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0

  So v = 12.58 m/s or v = -15.58 m/s ( not possible)

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7 0
2 years ago
Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate
Wittaler [7]

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J

Time required to burn the fat

t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s

Distance traveled by the bird

s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m

The bird will fly 455165.278 m

4 0
2 years ago
Read 2 more answers
One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the
sergeinik [125]
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4 0
3 years ago
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A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b
Nina [5.8K]

Answer: SG = 2.67

Specific gravity of the sand is 2.67

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Vs = 37.5cm^3

Density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

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Therefore, the specific gravity of sand is

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

Specific gravity of the sand is 2.67

3 0
2 years ago
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