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ch4aika [34]
3 years ago
11

What is the outcome of a star that runs out of hydrogen?

Physics
2 answers:
Vika [28.1K]3 years ago
6 0
These stars fuse helium into carbon just like the sun.
AlexFokin [52]3 years ago
3 0
These stars fuse helium into carbon just like the sun.
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Which electrode in Thomson’s vacuum tube was positively charged
Taya2010 [7]

Anode

Explanation:

The anode in the gas discharge tube used by Thomson in his experiment was the positively charged electrode.

Using the gas discharge tube, Thomson made the remarkable discovery of cathode rays.

The rays moves from the negatively charged cathode to the positively charged anode. This indicated that the rays carry positive charges.

Some parts of the tube are:

  • Cathode - negatively charged electrode
  • Power source
  • Gas at low pressure
  • Outlet to vacuum pump

Learn more:

cathode brainly.com/question/12747250

#learnwithBrainly

8 0
3 years ago
Helpp!!! Will mark brainlst
masya89 [10]

Answer:

i dont know

Explanation:

cuz i dont know what it wrote its too blurry

5 0
3 years ago
3. Your friend says your body is made up of more than 99.9999% empty space. What do you think?
Nesterboy [21]

Answer:

I would agree with the statement. it's not just the body, but everything that we see is almost 99.9999% empty space

8 0
3 years ago
A scientist who wants to study the affects of fertilizer on plants sets up an experiment. Plant A gets no fertilizer, Plant B ge
Solnce55 [7]

Answer: Plant A

Explanation: The control group is participants who do not receive the experimental treatment.

8 0
3 years ago
The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi
Keith_Richards [23]

Answer:

Explanation:

  a )

from lens makers formula

\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}

\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

 

5 0
3 years ago
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