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iogann1982 [59]
3 years ago
6

A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. However, he is unable to lift the suit

case 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
1 answer:
kvasek [131]3 years ago
4 0

Answer:

b)

Explanation:

Normal force, is always directed upward the surface over which is placed the object, and can adopt any value, as required to meet Newton's 2nd Law.

In this case, as the external force on the suitcase pulls upward, in order  to counteract the influence of gravity, normal force is less than the weight of the suitcase, as follows:

F + Fn = m*g

⇒ Fn = m*g - F

So, the normal force is equal to the magnitude of the weight of the suitcase (m*g) minus the magnitude of the force of the pull (F) which is the same expressed by the statement b.

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The quantity of heat must be removed is 1600 cal or 1,6 kcal.

<h3>Explanation : </h3>

From the question we will know if the condition of ice is at the latent point. So, the heat level not affect the temperature, but it can change the object existence. So, for the formula we can use.

\boxed {\bold {Q = m \times L}}

If :

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<h3>Steps : </h3>

If :

  • m = mass of water = 20 g => its easier if we use kal/g°C
  • L = latent coefficient = 80 cal/g

Q = ... ?

Answer :

Q = m \times L \\ Q = 20 \times 80 = 1600 \: cal

So, the quantity of heat must be removed is 1600 cal or 1,6 kcal.

<u>Subject : Physics </u>

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3 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
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Answer:

T = 16.72 N

Explanation:

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At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

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