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4 years ago

14

Magnetism is produced by the motion of electrons as they A. move around the nucleus. B. spin on their axes. C. both A and B D. n

one of the above

2 answers:

kupik [55]4 years ago

7 0

c)

both A and B is correct

DerKrebs [107]4 years ago

6 0

I think that the answer is C

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A man weighing 750 n and a woman weighing 500 n have the same momentum. what is the ratio of the man's kinetic energy km to that

miss Akunina [59]

Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662

4 0

3 years ago

Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible

ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)

Case for Large binocular telescope:

$A_{mirror1} = \pi (\frac{8.4m}{2})^{2}$

$A_{mirror1} = 55.41m$

For the second mirror will be the same value

$A = A_{mirror1}+A_{mirror2}$

$A = 55.41m+55.41m$

$A= 110.82m$

Case for Keck 1 telescope:

$A = \pi (\frac{10m}{2})^{2}$

$A = 78.53m$

Case for Hobby-Ebberly telescope:

$A = \pi (\frac{9.2m}{2})^{2}$

$A = 66.47m$

Case for Subaru telescope:

$A = \pi (\frac{8.3m}{2})^{2}$

$A = 54.10m$

Case for Gemini North telescope:

$A = \pi (\frac{8m}{2})^{2}$

$A = 50.26m$

Case for Magellan 2 telescope:

$A = \pi (\frac{6.5m}{2})^{2}$

$A = 33.18m$

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Key term:</em>

<em>Photons: particles that constitute light. </em>

3 0

3 years ago

A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near sid

igor_vitrenko [27]

Answer:

a) attractiva, b) dF = $k \frac{Q_1 \ dQ_2}{dx}$ , c) F = $k Q_1 \frac{Q_2}{d \ (d+L)}$ , d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

dF = $k \frac{Q_1 \ dQ_2}{dx}$

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

∫ dF = $k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2$

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

λ = dQ₂ / dx

DQ₂ = λ dx

we substitute

F = $k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}$

F = k Q1 λ ( $-\frac{1}{x}$ )

we evaluate the integral

F = k Q₁ λ $(- \frac{1}{d+L} + \frac{1}{d} )$

F = k Q₁ λ $( \frac{L}{d \ (d+L)})$

we change the linear density by its value

λ = Q2 / L

F = $k Q_1 \frac{Q_2}{d \ (d+L)}$

d) we calculate the magnitude of F

F =9 10⁹ (-4.2 10⁻⁶) $\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}$

F = -1.09 N

the sign indicates that the force is attractive

3 0

3 years ago

An acrobat performs with a swing, as shown in the diagram . Which of these most likely happens as the acrobat moves from Point 1

UkoKoshka [18]

Acrobat part
A
Plants
b
Rocks
C

6 0

3 years ago

Read 2 more answers

¿Qué cantidad de calor se desprende cuando 125 g de vapor de agua a 140 ºC se enfrían y congelan produciendo 125 g de hielo a -1

tensa zangetsu [6.8K]

Answer:

i dont know

Explanation:

7 0

3 years ago