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nikdorinn [45]
3 years ago
10

Can someone write and short summary on how to hula hoop

Physics
2 answers:
velikii [3]3 years ago
5 0
You pick up the hula hoop and stand inside of it then pick it up will you’re inside it and hold to your waist and spin it then turn your hips
MissTica3 years ago
5 0

Answer:

First you take a hula hoop and hold it around your waist then you start moving your hips in a circle and let the hula  hoop go off your hands and you keep moving your hips. You keep moving your hips until you want to stop but it might fall. Just keep trying and you should get the hang of it in a few minutes.

Explanation:

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Shearing of the wool is done with special instruments called_____​
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Answer:

The machine used is called a squaring shear, power shear, or guillotine.

Explanation:

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You're holding an open house for Brenda. Walt and Mary walk in and ask you for a flyer that gives details about the property. At
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Walt and Mary are my Customers at this point.

To construct proper client relationships you need to: greet clients and approach them in a way that is herbal and suits the character scenario. show customers that you recognize what their desires are. be given that a few humans may not want your merchandise and concentrate on constructing relationships with people who do.

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8 0
1 year ago
An asteroid is on a collision course with Earth. An astronaut lands on the rock to bury explosive charges that will blow the ast
forsale [732]

Answer:

The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters

Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, h_{max} = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g·h_{max}

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, v_e ≈ 3.13 m/s is therefore;

3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}

r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

7 0
3 years ago
Hii please help i’ll give brainliest!!
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Answer:

C

Explanation:

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the answer is (A) the movement of the magnet relative to the coil

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