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3 years ago

GIVING BRAINLIEST PLEASE HELP!!

Physics

1 answer:

andrey2020 [161]3 years ago

4 0

Answer:

B is the answer!!

Explanation:

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Two point charges, q1 = 2.0 × 10−7 C and q2 = −6.0 × 10−8 C, are held 25.0 cm apart. (a) What is the electric field at a point 5

AlladinOne [14]

Answer:

a) $432000\frac{N}{C}\hat{i}$

b) $-6.92\times10^{-14}N\hat{i}$

Explanation:

The magnitude of the electric field generated by a charged particle at a distance r is:

$E= k\frac{|Q|}{r^{2}}$

With Q the charge of the particle and k the constant ()

So, the electric field generated by q1 knowing that the point 5.0 cm apart the negative charge is $25.0cm-5.0cm=20.0 cm=0.2m$ apart the positive charge is:

$E_1= k\frac{|q1|}{r_1^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|2.0\times10^{-7}|}{(0.2)^{2}}$

$E_1= 45000\frac{N}{C}$

and the electric field generated by q2:

$E_2= k\frac{|q2|}{r_2^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|-6.0\times10^{-8}|}{(0.05)^{2}}$

$E_2=216000\frac{N}{C}$

Those are the magnitudes of the electric field, but electric field is a vector quantity, so the direction is important. Electric field generated by negative particles points towards the charge and electric field generated by positive particles points away the particle. So, if we define positive direction towards negative particle (x-axis):

$\overrightarrow{E_2}=+216000\frac{N}{C}\hat{i}$

$\overrightarrow{E_1}= +45000\frac{N}{C}\hat{i}$

Vector quantities satisfy superposition principle, this is $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$ , with E the total electric field.

$\overrightarrow{E}=(216000+45000)\hat{i}=432000\frac{N}{C}\hat{i}$

b) The force is:

$\overrightarrow{F}=e*\overrightarrow{E}$ ,

with q the charge of an electron

$\overrightarrow{F}=(-1.61\times10^{-19})*(432000)\hat{i}=-6.92\times10^{-14}N\hat{i}$

8 0

3 years ago

A skull believed to belong to an ancient human being has a carbon-14 decay rate of 5.4 disintegrations per minute per gram of ca

larisa86 [58]

Answer:

9.43*10^3 year

Explanation:

For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation

To start with, we use the formula

t(half) = In 2/k,

if we make k the subject of formula, we have

k = in 2/t(half), now we substitute for the values

k = in 2 / 5730

k = 1.21*10^-4 yr^-1

In(A/A•) = -kt, on rearranging, we find out that

t = -1/k * In(A/A•)

The next step is to substitite the values for each into the equation, giving us

t = -1/1.21*10^-4 * In(5.4/15.3)

t = -1/1.21*10^-4 * -1.1041

t = 0.943*10^4 year

8 0

4 years ago

Question 7 of 25

tiny-mole [99]

Answer:

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

2Na + MgCl₂ → 2NaCl + Mg

Explanation:

A balanced chemical equation is a chemical equation that have an equal number of elements of each type on both sides of the equation

Among the given chemical reactions, we have;

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

In the above reaction;

The number of phosphorus, P, on either side of the equation = 2

The number of bromine atoms, Br, on either side of the equation = 6

The number of chlorine atoms, Cl, on either side of the equation = 6

Therefore, the number of elements in the reactant side and products side of the reaction are equal and the reaction is balanced

The second balanced chemical reaction is 2Na + MgCl₂ → 2NaCl + Mg

In the above reaction, there are two sodium atoms, Na, one magnesium atom and two chlorine atoms on both sides of the reaction, therefore, the reaction is balanced

6 0

3 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

A helium balloon is filled to a volume of 2.88 x 103 L at 722 mm Hg and 19Â°C. When the balloon is released, it rises to an alti

inn [45]

Answer:

V₂=4.57 x 10³ L

Explanation:

Given that

V₁= 2.88 x 10³ L

P₁=722 mm Hg

T₁ = 19°C

T₁ =292 K

P₂=339 mm Hg

T₂= - 55°C

T₂=218 K

Lets take final volume = V₂

We know that ideal gas equation

PV = m R T

By applying mass conservation

$\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

$\dfrac{722\times 2.88\times 10^3}{292}=\dfrac{339\times V_2}{218}$

V₂=4.57 x 10³ L

Therefore volume will be 4.57 x 10³ L

3 0

3 years ago