Which statement describes the formation of the Milky Way galaxy?

What is the magnitude (in N/C) and direction of an electric field that exerts a 4.00 x 10^−5 N upward force on a −2.00 µC charge

emmasim [6.3K]

Answer:

The magnitude of electric field is $1.25\times10^{14}\ N/C$ in downward.

Explanation:

Given that,

Force $F= 4.00\times10^{-5}\ N$

Charge q= -2.00 μC

We know that,

Charge is negative, then the electric field in the opposite direction of the exerted force.

We need to calculate the magnitude of electric field

Using formula of electric force

$F = qE$

$E = \dfrac{F}{q}$

$E=\dfrac{4.00\times10^{-5}}{-2.00\times1.6\times10^{-19}}$

$E=-1.25\times10^{14}\ N/C$

Negative sign shows the opposite direction of electric force.

Hence, The magnitude of electric field is $1.25\times10^{14}\ N/C$ in downward.

8 0

4 years ago

(a) A standard sheet of paper measures 8 1/2 by 13 inches. Find the area of one such sheet of paper in m2.8.5(!meter/39.37in) —

Sergeu [11.5K]

Answer:

a) A = 0.07129 m²

b) A / A ’= 1.77

Explanation:

In this exercise we are asked to find the area in SI units, so let's start by reducing the dimensions to SI units.

width a = 8.5 inch (2.54 cm 1 inch) (1 m / 100 cm)

a = 0.2159 m

length l = 13 inch (2.54 cm / 1 inch) (1 m / 100 cm)

l = 0.3302 m

The area of a rectangle is

A = l a

A = 0.3302 0.2159

A = 0.07129 m²

b) we have a second sheet with reduced dimensions

a ’= 3/4 a

l ’= ¾ l

Let's find the area of this glossy sheet

A ’= l’ a ’

A ’= ¾ l ¾ a

A ’= 9/16 l a

To find the factor we divide the two quantities

A / A ’= l a 16 / (9 l a

A / A ’= 1.77

8 0

3 years ago

For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equat

zvonat [6]

Answer:

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

Explanation:

The Rydberg equation is given by

$\frac{1}{\lambda} =\frac{2\pi mk^2e^4}{h^3c} t (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

m is the mass of electron

k = 1/4π∈₀

∈₀ = is the permitivity of free space

e is the charge of electron

h is the plank constant

c is the speed of light in vaccum

z is the atomic number = 1

$\frac{1}{\lambda} =R (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

where R is the Rydberg constant = 1.097373 × 10⁷m⁻¹

For Paschen series of H spectrum

$n_f = 3$

$n_i = 5,6,7 ...$

in Paschen series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{16} )\\\\\lambda_m_a_x=1874.6nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=820.14nm$

The brackett series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{25} )\\\\\lambda_m_a_x=4050.05nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=1458.03nm$

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

8 0

3 years ago

A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

Nostrana [21]

Explanation:

Gauss Law relates the distribution of electric charge to the resulting electric field.

Applying Gauss's Law,

EA = Q / ε₀

Where:

E is the magnitude of the electric field,

A is the cross-sectional area of the conducting sphere,

Q is the positive charge

ε₀ is the permittivity

We be considering cases for the specified regions.

Case 1: When r < R

The electric field is zero, since the enclosed charge is equal to zero

E(r) = 0

Case 2: When R < r < 2R

The enclosed charge equals to Q, then the electric field equals;

E(4πr²) = Q / ε₀

E = Q / 4πε₀r²

E = KQ /r²

Constant K = 1 / 4πε₀ = 9.0 × 10⁹ Nm²/C²

Case 3: When r > 2R

The enclosed charge equals to Q, then the electric field equals;

E(4πr²) = 2Q / ε₀

E = 2Q / 4πε₀r²

E = 2KQ /r²

4 0

3 years ago

A gas is enclosed in a confainer fitted with a piston of cross sectional area 0.10 the pressureof the gas is maintained in 8000

Arlecino [84]

Answer:

10 Joule

Explanation:

The solution and answer are well written in the Pic above.

6 0

3 years ago