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Vika [28.1K]
3 years ago
8

What is the total net charge of an atom that contains 13 protons and 10 electrons? *

Physics
1 answer:
umka2103 [35]3 years ago
4 0

Answer:

It should be *+3

Explanation:

Al3+ indicates an ion of aluminum having a charge of + 3. I.e., since an aluminum atom normally has 13 protons and 13 electrons, this ion has 10 electrons (-10 charge) and 13 protons (+ 13 charge) giving it a charge of + 3 (-10 + 13 = +3).

The other i.d.i.ot answered unhelpfully

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Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate
vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>

5 0
3 years ago
What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

Solving for v gives

v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

6 0
4 years ago
A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it.
Juli2301 [7.4K]

Answer:

Explanation:

The change in force = 9.81(1.95 - 0.300) = 16.2 N

The change in length is 0.750 - 0.200 = 0.550 m

K = ΔF/Δx = 16.2/0.550 = 29.4 N/m

5 0
2 years ago
On a frictionless plane inclined at angle 303. The box is connected via a cord of negligible mass to a box of laundered money (m
KATRIN_1 [288]

Answer:

 T = 1.73 kg

Explanation:

Let's use Newton's second law for this balance problem

We draw a coordinate system with the x axis parallel to the plane and the Y axis perpendicular

The only force we have to lay down is the weight (W)

       Wx = W sin θ

       Wy = W cos θ

       Wx = 2.0 ain 60  

       Wx = 1.73 kg

       Wy = 2.0 cos 60

       Wy = 1.0 kg

Y Axis

      N-Wy = 0

X axis

      T- Wx = 0

      T = Wx

      T = 1.73 kg

7 0
3 years ago
Easy quiz easy points 7
Darya [45]
The answer is 2: hush
5 0
3 years ago
Read 2 more answers
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