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Svetach [21]
3 years ago
12

Why do many people view controls negatively?

Physics
1 answer:
GREYUIT [131]3 years ago
5 0
Many People<span> Are Averse to Management "</span>Control<span>" ... and then subsequent decisions about what to </span>do<span> is the essence of management coordination</span>
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Atoms of the gas neon
Alexxandr [17]

Explanation:

Neon is an atom with atomic number ten. Its atomic weight is 20.179 which cause it to have ten neutrons and ten protons in its nucleus and ten electrons outside. Neon; Neon, Ne, is a colorless inert noble gas and it is also the second lightest noble gas.

6 0
3 years ago
A 3.00-kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determ
aniked [119]

Answer:

a) k= 3594,7 N/m

b) v= 0.55 m/s

Explanation:

  • a)
  • As the surface is horizontal, the only change in energy will be the change in kinetic energy, as the box comes to an stop after compressing the spring.
  • As we know that the surface is frictionless also, this change in kinetic energy must be equal to the change in the elastic potential energy of the spring.
  • So we can write the following equality:

       \Delta K = \Delta U

       where \Delta K = \frac{1}{2}*m*v^{2}

       and \Delta U = \frac{1}{2} * k* \Delta x^{2}

  • Simplifying and replacing by the values, we get:

        3.00 kg* (1.8 m/s)^{2} = k* (0.052 m) ^{2}        

  • Solving for k:

k = \frac{3.00kg*(1.8m/s)^{2} }{(0.052m)^{2}} = 3594.7 N/m

  • k = 3594.7 N/m
  • b)
  • For this part, we can just apply the same equality, replacing the value of k by the one we got, and solving for the initial speed v:

        v = \sqrt{\frac{k*\Delta x^{2}}{m} } = \sqrt{\frac{3594.7N/m*(0.016m)^{2} }{3.00kg}} = 0.55 m/s

  • v = 0.55 m/s
6 0
3 years ago
A simple generator is used to generate a peak output voltage of 19.0 V . The square armature consists of windings that are 6.65
salantis [7]

One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,

\epsilon = NBA\omega

Where,

N = Number of loops

B = Magnetic Field

A = Cross-sectional Area

\omega = Angular velocity

Re-arrange to find N,

N = \frac{\epsilon}{BA\omega}

Our values are given as,

\epsilon = 19V

B = 0.434T

\omega = 49.8\frac{rev}{s} (\frac{2\pi rad}{1 rev}) = 99.6\pi rad/s

A = (6.65*10^{-2})^2 m^2

Replacing at our equation we have:

N = \frac{\epsilon}{( 0.434)A\omega}

N = \frac{19}{( 0.434)((6.65*10^{-2})^2)(99.6\pi)}

N = 31.63 \approx 32

Therefore the number of loops of wire should be wound on the square armature is 32 loops

6 0
3 years ago
A 210 Ohm resistor uses 9.28 W of
IgorLugansk [536]
  • Resistance=R=210Ohm
  • Power=9.28W=P

Current=I

\boxed{\sf P=I^2R}

\\ \sf\longmapsto I^2=\dfrac{P}{R}

\\ \sf\longmapsto I^2=\dfrac{9.28}{210}

\\ \sf\longmapsto I^2\approx0.04

\\ \sf\longmapsto I\approx\sqrt{0.04}

\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}

\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}

\\ \sf\longmapsto I\approx\dfrac{2}{10}

\\ \sf\longmapsto I\approx0.2A

4 0
2 years ago
At a wedding reception, you notice a child who looks like their mass is about 25 kg running across the dance floor then sliding
Vitek1552 [10]

Mass of the object m = 25 kg

Coefficient of friction Uk = 0.15

Frictional force Ff = Uk x F => Ff = Uk x m x g

Ff = 0.15 x 25 x 9.8

Frictional Force Ff = 36.75 N

4 0
3 years ago
Read 2 more answers
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