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Pavlova-9 [17]
3 years ago
6

What are the three stances in golf?

Physics
2 answers:
VARVARA [1.3K]3 years ago
8 0

Answer:

A golf stance is made up of several elements – the right width of stance, the right amount of knee flex, the angle of the feet and overall posture. We'll look at posture in part 6 of this series. A stance that is too narrow will prevent you from generating a lot of power in your golf swing.

Explanation:

Wewaii [24]3 years ago
4 0

Answer:

P1: Address.  

P2: Club Shaft Parallel With Ground On Takeaway.

P3: Lead Arm Parallel With Ground.

The center of your upper spine, your knees, and balls of your feet should line up on top of each other.

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A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

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How are Earth and Venus similar? How is Venus different from Earth? (Provide two similarities and two differences.)
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The movement of water into the atmosphere as it changes from a liquid to a gas is called _____.
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The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
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