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Trava [24]
2 years ago
15

An astronomer discovers a new galaxy using a telescope. The astronomer wants to investigate how the galaxy is moving relative to

the Milky Way galaxy.
In one or two sentences, make a hypothesis about the movement of the galaxy and explain at least one way to test the hypothesis.
Physics
1 answer:
belka [17]2 years ago
5 0

No matter what, if you’re in an already existing galaxy, any galaxy outside of your existing territory will be moving away from yours.

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Two balls of mass 0.09 kg hang on strings attached to the same point on the ceiling. The balls are given charges Q that cause th
telo118 [61]

Answer:

Q = 6.33μC

Explanation:

To find the value of the charge Q you take into account both gravitational force and electric force over each ball. By symmetry you can use the fact that both balls experiences the same forces. Hence you only take into account the forces for one ball for the x component and y component:

-Mg+Tcos\theta=0\\\\F_e-Tsin\theta=0

M: mass of the ball = 0.09kg

T: tension of the string

F_e: electric force between charges

angle = 45°

The electric force is given by:

F_e=k\frac{Q^2}{r^2}

Q: charge of the balls

r: distance between balls = 2m

You divide both equation in order to eliminate the tension T:

tan\theta=\frac{F_e}{Mg}=k\frac{Q^2}{Mgr^2}

By doing Q the subject of the formula and replacing you obtain:

Q=\sqrt{\frac{tan\theta Mgr^2}{k}}=\sqrt{\frac{tan45\°(0.09kg)(2m)^2}{(8.89*10^{9}Nm^2/C^2)}}=6.33*10^{-6}C=6.33\mu C

hence, the charge of the balls is 6.33μC

4 0
3 years ago
A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washe
STALIN [3.7K]

To solve this exercise it is necessary to take into account the concepts related to thermal expansion.

The thermal expansion is given by the function,

\Delta L = L_0 \alpha \Delta T

Where,

\Delta L= Change in Length

\Delta T =Change in Temperature

\alpha =Coeficiente de dilatación lineal

L_0 = Initial Length

By quickly deducing the formula, we can realize that the greater the change in temperature, the greater the change in the length of the radius.

The change in length is proportional to the change in temperature. Considering that the other two terms are constant we have that the correct one would be: <em>The hole in the center of the washer will expand.</em>

4 0
3 years ago
Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0
2 years ago
When touched by a plastic straw, the metal sphere will do what?
bearhunter [10]
I believe that if you touch a metal sphere with a plastic straw, the straw would not have enough strength to push it. So in that case, the metal sphere will not move and will stay in one place.

Please rate a 5 star
5 0
3 years ago
When you stretch a spring 20 cm past its natural length, it exerts a force of 8
Nastasia [14]

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

6 0
3 years ago
Read 2 more answers
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