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FrozenT [24]
3 years ago
10

An ant does 1 N-m of work in dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar.

Physics
1 answer:
algol133 years ago
8 0
The ant would drag the sugar 500m


Good luck!
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An 8 foot metal guy wire is attached to a broken stop sign to secure its position until repairs can be made. Attached to a stake
Angelina_Jolie [31]

Answer:

Explanation:

The guy wire is making a right angled triangle  with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.

base / hypotenuse = cos51

base = hypotenuse x cos51

= 8 x cos51

= 5.03 ft .

The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .

7 0
3 years ago
When the hydrogen atom makes the transition from the n=2 to the n=1 energy level, it emits a photon. This photon can be absorbed
bazaltina [42]

Answer:

n_{fn}= 4

Explanation:

To solve this exercise we will use Bohr's atomic model

               E_{n} = - 13.606 / n²     [eV]

The transition from level n = 2 to level n = 1 is valid

               E_{21} = - 13.606 [¼ -1/1]

               E_{21} = 10.2045 eV

Bohr's model for atoms with only one electron is

               E_{n} = -13.606 Z² / n²

Where Z is the atomic number of the atom.

In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches

         ΔE = -13.606 [4 /  n_{fn}² - 4/4]

         4 / n_{fn}² = -ΔE / 13.606 + 1

         4 / n_{fn}² = -10.2045 / 13.606 +1 = -0.75 +1

         4 / n_{fn}² = 0.25

        n_{fn} = √ 4 / 0.25

        n_{fn}= 4

8 0
3 years ago
A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt
anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

6 0
3 years ago
What two major uses does this (H-R)diagram have for astronomers?
EastWind [94]
66666666666666666666666666666666666666666666666666666666666666666666666666666666666666

7 0
3 years ago
Which of the following is not a way in which humans directly impact algal bloom production? a. industrial pollution b. wastewate
Monica [59]

d. fishing

The other options all directly affect algal bloom production because they affect the nutrients in the water, and an overabundance of certain nutrients in the water is what causes algal bloom.

6 0
3 years ago
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