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AVprozaik [17]
3 years ago
11

55N of force are applied onto a table, but the force of friction is 8N. What is the net force acting on the table? Please explai

n, I'm trying to learn how to actually do this
Physics
1 answer:
nirvana33 [79]3 years ago
8 0

Answer:

get a tutor or something to help man

Explanation:

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3 years ago
If stellar parallax can be measured to a precision of about 0.01 arcsec using telescopes on the Earth to observe stars, to what
marin [14]

Answer:

It corresponds to a distance of 100 parsecs away from Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

It is defined in a analytic way as it follows:

       

\tan{p} = \frac{1AU}{d}

Where d is the distance to the star.

p('') = \frac{1}{d} (1)  

Equation (1) can be rewritten in terms of d:

d(pc) = \frac{1}{p('')} (2)

Equation (2) represents the distance in a unit known as parsec (pc).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).

For the case of   (p('') = 0.01):

d(pc) = \frac{1}{0.01}

d(pc) = 100

Hence, it corresponds to a distance of 100 parsecs away from Earth.

<em>Summary:</em>

Notice how a small parallax angle means that the object is farther away.

Key terms:

Parsec: Parallax of arc second

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3 years ago
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3 years ago
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A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

4 -in.-thick  

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

                           = T/2 = 0.952/2 = 0.476 kips/in²

8 0
3 years ago
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