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Alekssandra [29.7K]

3 years ago

11

Why do atoms form bonds? A. to decrease the number of electron shells in an atom B. to increase the number of electrons orbiting

the nucleus C. to transfer electron shells between atoms D. to create a completely full outer shell of electrons

1 answer:

Natali [406]3 years ago

7 0

Its C to transfer electron shells between atoms

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10. Someone takes 11 minutes to walk up a hill 120m high. His weight is 550N.

belka [17]

Answer:

okay with you if you want to

8 0

2 years ago

If 2 grams of element X combine with 4 grams of element Y to form compound XY, how many grams of element Y would combine with 46

Butoxors [25]

92 grams of Y would combine with 46 grams of X

3 0

3 years ago

Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?

storchak [24]

Answer:

$\large \boxed{42\, \mu \text{C}}$$

Explanation:

The formula for the force exerted between two charges is

$F=k \dfrac{ q_1q_2}{r^2}$

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

$F=k\dfrac{q^{2}}{r^2}$

$\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}$

7 0

3 years ago

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

5 0

3 years ago

A skater of mass 40 kg is carrying a box of mass 5 kg. The skater has a speed of 5 m/s with respect to the floor and is gliding

Yuki888 [10]

For the part a) we need only the momentum of the box and we have the data to find it.

Momentum is given by,

$p=mv$

where clearly, p is the momentum, m the mass of the box and v is the velocity.

Substituting,

$p=(5kg)(5m/s)\\p=25kg.m/s$

For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,

$p=(5kg)(5m/s)\\p=25kg.m/s$

<em>It is the same that part a)</em>

8 0

2 years ago