How to make an uncharged particle positively charged

Barnard's star is a near neighbor of the Sun whose properties we know quite well. It is a type M4V with absolute magnitude 13.22

vitfil [10]

Answer:

The star is at a distance of 100 parsecs.

Explanation:

The distance can be determined by means of the distance modulus:

$M - m = 5log(d) - 5$ (1)

Where M is the absolute magnitude, m is the apparent magnitude and d is the distance in units of parsec.

Therefore, d can be isolated from equation 1

$log(d) = (M - m + 5)/5$

Then, Applying logarithmic properties it is gotten:

$d = 10^{(M - m + 5)/5}$ (2)

The absolute magnitude is the intrinsic brightness of a star, while the apparent magnitude is the apparent brightness that a star will appear to have as is seen from the Earth.

Since both have the same spectral type is absolute magnitude will be the same.

Finally, equation 2 can be used:

$d = 10^{(13.22 - 8.22+ 5)/5}$

$d = 100 pc$

Hence, the star is at a distance of 100 parsecs.

Key term:

Parsec: Parallax of arc seconds

8 0

3 years ago

PART ONE

Korvikt [17]

Answer:

1) 460.5 N

2) 431.7 N

Explanation:

Draw a free body diagram. There are four forces on the hammer:

Applied force 62.5 N in the +x direction, 30 cm from the ground

Reaction force Rᵧ in the +y direction, at the point of contact

Reaction force Rₓ in the +x direction, at the point of contact

Reaction force F at 31° from the vertical, 4.75 cm to the left of the point of contact

Part One

To find F, sum the moments about the point of contact:

∑τ = Iα

(62.5 N) (30 cm) − (F cos 31°) (4.75 cm) = 0

F = 460.5 N

Part Two

To find Rₓ and Rᵧ, sum the forces in the x and y directions.

∑Fₓ = ma

62.5 N − F sin 31° + Rₓ = 0

Rₓ = 174.7 N

∑Fᵧ = ma

-F cos 31° + Rᵧ = 0

Rᵧ = 394.7 N

The net reaction force at the point of contact is:

R = √(Rₓ² + Rᵧ²)

R = 431.7 N

7 0

3 years ago

I will Mark Brainliest 1. ) Allena and Charissa were discussing whether Cl2 is an element or a compound. Allena said that it is

WITCHER [35]

The answer to question one is A.
The answer to question two is A.
The answer to question three is D.

7 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

Two identical billiard balls traveling at the same speed have a head-on collision and rebound. If the balls had twice the mass,

Vinvika [58]

Answer:

a) No difference

Explanation:

Since the billiard balls are identical , they have the same mass. Also they have the same speed

Since the angular momentum is conserved and the total energy is conserved ( if we assume elastic collision)

1/2 m1 * v i1² +1/2 m2 * v i1² = 1/2 m1 * v f1² +1/2 m2 * v f2²

where m= mass , vi= initial velocity , vf= final velocity

since m1=m2=m , vi1=vi2=vi

1/2 m1 * v i1² +1/2 m2 * v i1² = 1/2 m1 * v f1² +1/2 m2 * v f2²

m * v i² = 1/2 m (v f1² +v f2² )

vi² = 1/2(v f1² +v f2² )

since the 2 balls are indistinguishable from each other (they have identical initial mass and velocity) there is no reason for a preferential speed for one of the balls and therefore its velocities must be equal . Thus vf1=vf2=vf

therefore

v i² = 1/2(v f1² +v f2² ) = v i1² = 1/2* 2vf² = vf²

and thus

vi= vf

in conclusion, there is no difference in speed after the rebound

4 0

3 years ago