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Elden [556K]
3 years ago
15

How many board feet of lumber are needed assuming 200 lin. Ft. of 2""x 6"" lumber a. 12 b. 16.67 c. 120 d. 200

Physics
1 answer:
Svetach [21]3 years ago
4 0

Answer:

B) 16.67

Explanation:

If the dimension of one lumber is 2" × 6", the total area of one lumber will be 12inch²

If the total board feet of lumber there is 200in, therefore the total board of lumber that will be needed is 200/12 which gives 16.67 lumbers

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If an asteroid is 4 AU from the Sun, what is the period of revolution around the Sun for the asteroid?
Vedmedyk [2.9K]

Answer: 8 years

Explanation:

According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”:</em>

<em />

T^{2}\propto a^{3} (1)

In other words: this law states a relation between the orbital period T of a body (moon, planet, satellite, comet, asteroid) orbiting a greater body in space (the Sun, for example) with the size a of its orbit.  

However, if T is measured in years (Earth years), and a is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:

T^{2}=a^{3} (2)

This means that now both sides of the equation are equal.

Knowing a=4 AU and isolating T from (2):

T=\sqrt{a^{3}} (3)

T=\sqrt{(4 AU)^{3}} (4)

Finally:

T= 8 yearsThis is the period of the asteroid

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Knowing the definition of work as used by physicists, how can the work done on an object be equal to zero? a. The force and dist
Nuetrik [128]

Answer:

C

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Explain why the direction of the south equatorial current changes
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Some scientist believe that the ozone layers of the earth had been weakening and the waters or current changes direct ever 5-7 month.
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A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it
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Answer:

14 m/s

Explanation:

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Solving for v, we find

v=\sqrt{u^2+2gh}=\sqrt{0+2(9.8 m/s^2)(10 m)}=14 m/s

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Hey! How's it going? If you need anything, feel free to send me a friend request and message me.

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