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kolezko [41]
3 years ago
14

The amp is the unit for _________.

Physics
2 answers:
adell [148]3 years ago
7 0
B.) <span>The amp is the unit for "Current"

Hope this helps!</span>
Charra [1.4K]3 years ago
3 0
The ampere is the unit of current, and it's symbol is 'A' 

hope this helped
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D. When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.

Explanation:

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A pole-vaulter converts the kinetic energy of running to elastic potential energy in the pole, which is then converted to gravit
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A particle hangs from a spring and oscillates with a period of 0.2s.If the mass-spring system remained at rest, by how much woul
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Answer:

The distance the mass would stretch it is    x = 0.00992948

The correct option is A

Explanation:

From the question we are told that

           The period of the slit is T = 0.2s

           The acceleration due to gravity is g =9.8 m/s^2

Generally the period is mathematically represented as

                     T = 2 \pi \sqrt{\frac{m}{k} }

          Whee m is the particle and k is the spring constant

        making \frac{m}{k} the subject

                        \frac{m}{k}  = [\frac{T}{2 \pi} ]^2

The weight on the particle is related to the force stretching the spring by this mathematical relation

               W = F_s

              mg = kx

where x is the length by which the spring is stretched

          \frac{m}{k}  = \frac{x}{g}

Substituting the equation above for \frac{m}{k}

            [\frac{T}{2 \pi} ]^2 = \frac{x}{g}

making x the subject

              x = g [\frac{T}{2 \pi} ]

Substituting the value

            x = 9.8 * [\frac{0.2}{2 * 3.142} ]^2

            x = 0.00992948

6 0
3 years ago
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You cause a particle to move from point A, where the electric potential is 11.3 V, to point B, where the electric potential is −
BabaBlast [244]

Explanation:

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Using this definition, we can calculate the electrostatic potential energy change between point A and B:

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\Delta U=1.6*10^{-19}C(11.3V+25.9V)\\\Delta U=5.952*10^{-18}J

Neutral hydrogen atom: q=0

\Delta U=0

Singly ionized helium atom: q=1.6*10^{-19}C

\Delta U=1.6*10^{-19}C(11.3V+25.9V)\\\Delta U=5.952*10^{-18}J

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Answer:

C) 20

Explanation:

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