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Anuta_ua [19.1K]
3 years ago
13

A +0.0129 C charge feels a 4110 N

Physics
1 answer:
Savatey [412]3 years ago
4 0

Answer:

r = 14.13 m

Explanation:

Given that,

Charge 1, q₁ = +0.0129 C

Charge 2, q₂ = -0.00707 C

The force between charges, F = 4110 N

We need to find the distance between charges. The formula for the force between charges is given by :

F=k\dfrac{q_1q_2}{r^2}

Where

r is the distance between charges

So,

r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 0.0129 \times 0.00707 }{4110 }} \\\\r=14.13\ m

So, the distance between charges is equal to 14.13 m.

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Feliz [49]

Answer:

Beryllium bromide

Explanation:

There exists 2 elements in the question

Beryllium Be with valence of +2, and it's a metal

Bromine Br with a valence of -1, and is a non - metal.

Both of them combine together forming an ionic bond:

Be + Br(2) -> BeBr(2)

In naming the compound, we would say that since the first is a metal, beryllium, and the second being a halogen, bromine. We name it by calling the first name together with the last name and changing the last name of the non metal from -ine to -ide.

So we have, Beryllium Bromide.

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3 years ago
a vehicle changes its velocity from 90 km/h to a dead stop in 10s. show that its acceleration in doing so is -2.5m/s^2
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90000m/3600s = 25 m/s

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A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from it
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Answer:

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Microwave ovens emit microwave energy with a wavelength of 12.3 cm. What is the energy of exactly one photon of this microwave r
Valentin [98]
The energy of a photon is proportional to the speed of light and inversely proportional to the wavelength. The constant of proportionality is Planck's constant, <span>6.626 * 10^-34 Js</span>

E = hc/ {wavelength} 
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4 0
3 years ago
Read 2 more answers
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
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