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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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The planet Gallifrey has 2 times the gravitational field strength and 2 times the radius of the Earth.
Alexeev081 [22]

The mass of the planet Gallifrey is 8 times the mass of the Earth.

  • let the gravitational field of Earth = g
  • let the radius of the Earth = R
  • gravitational field of Gallifrey = 2g
  • radius of Gallifrey = 2R

<h3>What is gravitational potential energy?</h3>
  • This is the work done in moving an object to a certain distance against gravitational field.

The gravitational field strength of the Earth is given as follows;

g = \frac{GM}{r^2} \\\\G = \frac{gr^2}{M}

The gravitational field strength of the Planet Gallifrey is calculated as follows;

g_2 = \frac{GM_2}{r_2^2}

G = \frac{g_2r_2^2}{M_2} \\\\\frac{g_2r_2^2}{M_2} = \frac{gr^2}{M}\\\\M_2gr^2 = Mg_2r_2^2\\\\M_2 = \frac{Mg_2r_2^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times (2r)^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times 4r^2}{gr^2} \\\\M_2 = 8M

Thus, the mass of the planet Gallifrey is 8 times the mass of the Earth.

Learn more about gravitational field strength here:  brainly.com/question/14080810

4 0
2 years ago
one layer of earth's atmosphere is called the stratosphere. At one point above earth's surface the stratosphere extends from an
masya89 [10]

According to all the given requirements 

stratosphere x starts at an altitude of 16 km

that means  

<span>16≤x

</span> and,as given ,stratosphere ends at an altitude of 30 km:
which means
<span>x≤30</span> combining both of them we get

<span>16≤x≤30</span> So the range of the stratosphere is from 16 to 30 km. 

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2 years ago
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Globalization concerns many workers because ________.
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Globalization concerns many workers because companies will tend to seek lower costs by outsourcing labor.
6 0
3 years ago
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A ball is thrown down vertically with an initial speed of v0 from a height of h. (a) What is its speed just before it strikes th
Alexus [3.1K]

Answer:

a)   v² = v₀² + 2 g h,  b)   t = v₀/g  (1+ √ (1 + 2gh/ v₀²))

Explanation:

a) This is an exercise that we can solve using conservation of energy.

Starting point. High point

         Em₀ = K + U = ½ m v₀² + m gh

Final point. Soil

         Em_{f} = K = ½ m v²

energy is conserved because there is no friction

         Emo = Em_{f}

         ½ m v₀² + m g h = ½ m v²

         v² = v₀² + 2 g h

b) the time it takes to reach the ground can be calculated with kinematics

let's create a reference frame with positive upward direction

         v = vo - g t

when it reaches the ground it has a velocity v, the initial velocity is downwards v₀ = -v₀

        v = -v₀ - gt

        t = - (v + v₀) / g

we substitute the velocity values ​​calculated in the previous part

        t = - (√(v₀² + 2 g h) + vo) / g

we will simplify the equation a bit

        t = - v₀/g  (1+ √ (1 + 2gh/ v₀²))

c) is now thrown vertically upward with the same initial velocity vo.

   To find the final velocity we use the conservation of energy where the velocity is squared, so it does not matter if it is positive or negative, therefore in this section the value should be the same as in part a

         v = √ (v₀² + 2gh)

d) for this part if there is change since the speed is not squared

     v₀ = v₀

          v = v₀ - gt

          t = (v₀ - v) / g

          t = (v₀ - √(v₀² + 2 g h)) / g

          t = v₀/g   (1 - √(1 + 2gh / v₀²))

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Yes the temp should be changed
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