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kap26 [50]
3 years ago
13

In a water molecule, which atom has the strongest attraction for shared electrons?

Physics
2 answers:
ValentinkaMS [17]3 years ago
7 0

B.  

oxygen for the questing above


ra1l [238]3 years ago
6 0
Oxygen... Hope this Warner helps
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How did you manage to overcome the difficulties that arose?
Flauer [41]

Answer:

I just toughed it out and talked with friends

Explanation:

4 0
2 years ago
How many light years away is the sun from the middle of the Millky way​
julsineya [31]

Answer:

The Milky Way is about 1,000,000,000,000,000,000 km (about 100,000 light years or about 30 kpc) across. The Sun does not lie near the center of our Galaxy. It lies about 8 kpc from the center on what is known as the Orion Arm of the Milky Way

4 0
3 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
3 years ago
A 6.5 x 104 W engine exerts a constant force on of 5.5 x 103 N on a car, the resulting velocity is?
jek_recluse [69]

Answer:

12m/s

Explanation:

Given parameters:

Power = 6.5 x 10⁴W

Force  = 5.5 x 10³N

Unknown:

The resulting velocity  = ?

Solution:

 The velocity of a body is related to force and power using the expression below;

            Power  = Force x velocity

Insert the parameters and solve for velocity

         6.5 x 10⁴  = 5.5 x 10³ x velocity

          velocity  = \frac{6.5 x 10^{4} }{5.5 x 10^{3} }   = 12m/s

4 0
2 years ago
) The centre of a mass, m, is at a distance, r, from the centre of a larger mass M. Assuming there are no other masses present,
ivanzaharov [21]

Answer:

U = - G m M / r

Explanation:

The gravitational potential energy is given by the expression

         U = - G m₁ m₂ / r

dodne G is the gravitational cosntnate (G = 6.67 10⁻¹¹¹), m and m are the mass of the bodies involved

subtype the given values

         U = - G m M / r

8 0
3 years ago
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