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4 years ago

I’m a little lost on this one need help

Physics

2 answers:

dsp734 years ago

8 0

The correct answer is the first options. If an atom is neutrally charged, it means it has the same number of protons (positive charge) as it does electrons (negative charge). A neutron is neither positive nor negative, it’s neutral, so the overall charge of the atom would remain neutral.

kari74 [83]4 years ago

6 0

False, because although neutrons have no charge, if you add more it will make the atom unbalanced in electric charges.

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A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.70 cm wide and 12.0 m long. Between the sheets is a mic

KIM [24]

Answer:

2.121876 mC

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $0.037\times 12\ m^2$

d = Thickness = 0.0225 mm

E = Dielectric strength = $1\times 10^8\ V/m$

k = Dielectric constant = 5.4

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}\\\Rightarrow C=\dfrac{5.4\times 8.85\times 10^{-12}\times 0.037\times 12}{0.0225\times 10^{-3}}\\\Rightarrow C=9.43056\times 10^{-7}\ F$

Maximum voltage is given by

$V_m=E_md\\\Rightarrow V_m=1\times 10^8\times 0.0225\times 10^{-3}\\\Rightarrow V_m=2250\ V$

Maximum charge is given by

$Q_m=CV_m\\\Rightarrow Q_m=9.43056\times 10^{-7}\times 2250\\\Rightarrow Q_m=0.002121876\ C=2.121876\ mC$

The maximum charge that can be stored in this capacitor is 2.121876 mC

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3 years ago

Identify the scientific law from among the following statements. A) The book fell 1 meter in 0.5 seconds B) Gravity acts upon ea

zalisa [80]

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3 years ago

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Your microwave will not turn on, and you speculate that a circuit breaker in the house has been tripped. In scientific terminolo

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Answer:

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3 years ago

When is the electric flux on a section of a closed surface positive?.

rodikova [14]

When the electric field pierces outward through the section.

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2 years ago

The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

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3 years ago