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3 years ago

If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?

Physics

1 answer:

storchak [24]3 years ago

4 0

Answer:

d.100 meters

Explanation:

The diameter of the Milky Way Galaxy is approximately 100,000 light years.

Here we are using 1 millimiter (1 mm) to represent 1 light-year (1 ly). So, we can set the following proportion:

$1 mm : 1 ly = x : 100,000 ly$

and by finding x, we find the diameter of the Milky Way Galaxy in the scale used:

$x=\frac{(1mm )(100,000 ly)}{1 ly}=100,000 mm = 100 m$

so the correct answer is

d. 100 meters

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The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 6 cm/s. When the length is

Fudgin [204]

Answer:

24cm/s

Explanation:

A=L*w

A'=L'*w'

L=13

w=5

L'=4

w'=6

A=?

A'=?

A=L*w

A=13*5

A=65

A'=L'*w'

A'=4*6

A'=24

*the given lengths are just to throw you off*

3 0

2 years ago

A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the

Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

Goodluck

Explanation:

4 0

2 years ago

A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant

mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0

2 years ago

What are the three ways of answering a scientific question

My name is Ann [436]

Answer:

Let's start by understanding what exactly a scientific question is. A scientific question is a question that may lead to a hypothesis and help us in answering (or figuring out) the reason for some observation. A good scientific question has certain characteristics. It should have some answers (real answers), should be testable.

Here's examples of a few:

Why is that a star?

What is that star made of?

Hope this can lead you to the answer you're looking for at least!!

5 0

2 years ago

The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

sattari [20]

Answer:

k = 11,564 N / m, w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

let's apply the equilibrium condition at this point

Axis y

W_{y} - Fr = 0

Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

sin 46 = $W_{y}$ / W

W_{y} = W sin 46

we substitute

mg sin 46 = k y

k = mg / y sin 46

If the length of the bar is L

sin 46 = y / L

y = L sin46

we substitute

k = mg / L sin 46 sin 46

k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

k = 1.18 9.8 / 1

k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

Em₀ = U = mgy

end point. Point at 30º

$Em_{f}$ = K -Ke = ½ I w² - ½ k y²

em₀ = Em_{f}

mgy = ½ I w² - ½ k y²

w = √ (mgy + ½ ky²) 2 / I

the height by 30º

sin 30 = y / L

y = L sin 30

y = 0.5 m

the moment of inertia of a bar that rotates at one end is

I = ⅓ mL 2

I = ½ 1.18 12

I = 0.3933 kg m²

let's calculate

w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

w = 6.06 rad / s

7 0

3 years ago