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3 years ago

Which of these best explains why people on Earth cannot see the entire shape of the Milky Way

1 answer:

5 0

I believe the answer is A.

Since the Earth is in the Milky Way and not outside it, we cannot see the exact shape of it. Physicists have been able to track and graph the movements of the planets accurately for thousands of years, but that does not mean we know the shape of the entire solar system.<span />

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A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = $4.5 \times 10^{-5} C$

Electric Field strength (E) = $2.0 \times 10^{4} N/C$

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

$U = q \times E \times d$

So now insert the values to find the answer

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

On further solving

$U = 0.027 J$

8 0

2 years ago

Which of these are part of our solar system? select all that apply

Len [333]

A)planets
b)the sun
c)moons
e)comets
f)asteroids

6 0

3 years ago

Read 2 more answers

How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n

grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

$dSin\theta = m\lambda$

Where

m= Any integer which represent the number of repetition of spectrum

$\lambda$ = Wavelength

d = Distance between the slits.

$\theta$ = Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

$d = \frac{m\lambda}{sin\theta }$

$d = \frac{2*536*10^{-9}}{sin(24)}$

$d = 2.63*10^{-6}m$

Therefore the number of lines per millimeter would be given as

$\frac{1}{d} = \frac{1}{2.63*10^{-6} }$

$\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})$

$\frac{1}{d} = 379.4 lines/mm$

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0

2 years ago

An object of mass 30 kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.

velikii [3]

Answer:

In free fall, mass is not relevant and there's no air resistance, so the acceleration the object is experimenting will be equal to the gravity exerted. If the object is falling on our planet, the value of gravity is approximately 9.81ms2 .

7 0

3 years ago

An object dropped from rest from the top of a tall building on planet x falls a distance d(t)18 left parenthesis t right parenth

melamori03 [73]

displacement is given by equation

$d = 18t^2$