<h2>Answer: polar covalent bonding</h2>
In polar covalent bonding, electrons are shared <u>unequally </u>between atoms, due to the unequal distribution of electrons between atoms of different elements, slightly positive and slightly negative charges appear in different parts of the molecule.
In this context, water molecule is an exellent example of this type of bonding:
Water (
) can stick to itself through hydrogen bonds, because a water molecule consists of 2 hydrogen atoms attached to 1 oxygen atom.
The oxygen atom tends to monopolize more electrons and keeps them away from hydrogen. Then, it can be said that a water molecule will have a negative side (oxygen) and a positive side (hydrogen).
Thanks to this polarity, water molecules can stick together with the formation of hydrogen bonds to attract a partial positive charge of hydrogen and a more electronegative atom, such as oxygen.
Answer:
magnitude of vector is 276.19 kg m/s
Explanation:
The initial momentum is vector of magnitude
And driven in a coherent manner with initial vector.
same magnitude is momentum after the impact, but it is oriented perpendicularly to initial momentum vector.
So, you have 2 momentum vector of specified magnitude perpendicular to one another.
The contrast between such two vectors is a right angle triangle hypotenuse of 195.3 sides
magnitude of vector is 
Answer:
= 30N - 10N = 20N
Please mark as brainliest!
It is appreciated hope this helps!
Explanation:
I have put what a simple free body force diagram that you would need for this question below!
Just swap out the 5N and 3N with 30N and 10N :)
{Answer}
A- (10kg sphere and 30m)
{I hope this helps :) }
Answer:
(a) 277.05 Ω
(b) 15.39 Ω
(c) 3.76 W
Explanation:
(a)
Applying,
P = V²/R.......................... Equation 1
Where P = Power dissipated by the string. V = Voltage source, R = equivalent resistance of the light string
Make R the subject of the equation
R = V²/P................... Equation 2
Given: V = 130, P = 61 W
Substitute into equation 2
R = 130²/61
R = 277.05 Ω
(b) The resistance of a single light is given as
R' = R/18 (since the light are connected in series and the are identical)
Where R' = Resistance of the single light.
R' = 277.05/18
R' = 15.39 Ω
(c)
Heat dissipated in a single light is given as
P' = I²R'..................... Equation 3
Where P' = heat dissipated in a single light, I = current flowing through each light.
We can calculate for I using
P = VI
make I the subject of the equation
I = P/V
I = 61/130
I = 0.469 A.
Also given: R' = 15.39 Ω
Substitute into equation 3
P' = 0.496²(15.39)
P' = 3.76 W
