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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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What needs to be the relation between the two forces to cause the upper magnet to float without moving?
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Answer:

If it points the other way, the fields subtract, for a lower energy, and so the magnet prefers to turn to point in this way. Magnets in uniform fields feel torques which make them turn around if they are not pointing in the right direction, but there is no net force making the magnet want to levitate.

Explanation:

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3 years ago
If an R = 1-kΩ resistor, a C = 1-μF capacitor, and an L = 0.2-H inductor are connected in series with a V = 150 sin (377t) volts
fgiga [73]

Answer

given,

R = 1-kΩ  = 1000 Ω

C = 1-μF

L = 0.2-H

V = V_max sin( ω t)

comparing

V = 150 sin ( 377 t)

ω = 377

\chi_c = \dfrac{1}{\omega C}

\chi_c = \dfrac{1}{377 \times 1 \times 10^{-6}}

\chi_c = 2652.5\Omega

\chi_L =377 \times 0.2

\chi_L =75.4\ \Omega

Impedance,

Z = \sqrt{R^2+(\chi_L-\chi_c)^2}

Z = \sqrt{1000^2+(75.4 -2652.5)^2}

Z = 2764.3 Ω

now,

V_{max} = 150 V

I_{max} = \dfrac{V}{Z}

I_{max} = \dfrac{150}{2764.3}

I_{max} = 0.0543

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3 0
3 years ago
TIMED! URGENT! REALLY APPRECIATE HELP!! TYSM!!!!!!
Diano4ka-milaya [45]
The answer is 27.03 I just multiplied the two numbers
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3 years ago
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Instead of rolling down the ramp over the step, imagine the ball falls off the back of the ramp directly to the ground. What is
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0J

According to the law of conservation of energy, the potential energy is converted to kinetic energy. Remember, potential energy is calculated using height and weight. If the ball is on the ground, height is 0.

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Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

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           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

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3 years ago
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