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3 years ago

9

How do variations in electronegativity result in the unequal sharing of electrons in polar molecules?

1 answer:

9966 [12]3 years ago

7 0

Answer:

Variations in electronegativity result in the unequal sharing of electrons in polar molecules because when one atom is more electronegative than the other, it becomes more polar than the other atom. ... Methane bonds are covalent because they share carbon and hydrogen valence electrons.xplanation:

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What will happen to this current if a magnet is brought near the cord? A. It will exert a force on the voltage. B. The electric

Andreyy89

Answer:

The correct answer is D)

Explanation:

When an electric magnet is brought near a cord with an electric current, the cord will most likely deflect away from the magnet because electric fields flowing through a wire generates its own magnetic field.

Cheers!

8 0

3 years ago

A car is a compound machine true or false

Viktor [21]

True is the answer. I believe

6 0

3 years ago

How would the composition of an atom change if both the atomic number and mass number each increase by one?

Softa [21]

If both the atomic number and mass number each increased by one, then ...

-- the atom would have one more proton in its nucleus,

and

-- whenever it was in a neutral state, it would also have one more electron in its cloud.

<em>choice - B</em>

7 0

4 years ago

If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in

Elza [17]

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, $N_P$ = 50 turns

number of turns in the secondary winding, $N_S$ = 10 turns

the secondary load resistance, $R_S$ = 250 Ω

Determine the turns ratio;

$K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5$

Now, determine the reflected resistance in the primary winding;

$\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms$

Therefore, the reflected resistance in the primary winding is 6250 Ω

6 0

3 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0 m/s at a 40.0 ∘ angle from the horizonta

Gre4nikov [31]

Answer:

Explanation:

given,

initial speed of the shot = 12.0 m/s

angle = 40°

height at which shot leaves her hand = 1.80 m

v_x = 12 cos 40° = 9.19 m/s

v_y = 12 sin 40° = 7.71 m/s

time to reach maximum height =

= $\dfrac{v_y}{g}$

= $\dfrac{7.71}{9.8}$

= 0.787 s

$h = v_y t + \dfrac{1}{2}gt^2$

h = 7.71 × 0.787 - 0.5 × 9.81 × 0.787²

h = 3.03 m

the maximum height attain = 3.03 + 1.8 = 4.83 m

now free fall from the maximum height

$h =\dfrac{1}{2}gt^2$

$4.83 = \dfrac{1}{2}\times 9.8 \times t^2$

t = 0.9928 s

total time = 0.9928 + 0.787 = 1.7798 s

range =

d = vₓ t

d = 16.36 m

8 0

4 years ago