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3 years ago

Which of the following explains why metallic bonding only occurs between

Physics

1 answer:

Y_Kistochka [10]3 years ago

3 0

Answer:

D. Metallic atoms have valence shells that are mostly empty, which

means these atoms are more likely to give up electrons and allow

them to move freely.

Explanation:

Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.

In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.

Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.

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A bicyclist, initially at rest, begins pedaling and gaining speed steadily for 4.90s during which she covers 25.0m.

emmasim [6.3K]

The bicyclist accelerates with magnitude a such that

25.0 m = 1/2 a (4.90 s)²

Solve for a :

a = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²

Then her final speed is v such that

v ² - 0² = 2a (25.0 m)

Solve for v :

v = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s

Convert to mph. If you know that 1 m ≈ 3.28 ft, then

(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h

8 0

2 years ago

Elina [12.6K]

Answer:

The vertical distance that the ski jumper fell is 417.45 m.

Explanation:

Given;

initial horizontal velocity of the jumper, $V_x$ = 26 m/s

horizontal distance of the jumper, dx = 240 m

The time of the motion is given by;

dx = Vₓt

t = dx / Vₓ

t = 240 / 26

t = 9.23 s

The vertical distance traveled by the diver is given by;

$d_y = V_yt + \frac{1}{2}gt^2$

initial vertical velocity, $V_y$ , = 0

$d_y = \frac{1}{2}gt^2\\\\d_y = \frac{1}{2}(9.8)(9.23)^2\\\\d_y = 417.45 \ m$

Therefore, the vertical distance that the ski jumper fell is 417.45 m.

6 0

3 years ago

Which of the following have the same density. number 26

marusya05 [52]

1 cubic cm is the same as 1 mL, so the answer would be C.

8 0

2 years ago

Consider three scenarios in which a particular box moves downward under the pull of gravity :

luda_lava [24]

Answer:

1. True WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

WA = W h cos 0

WA = mg h

B) On a ramp without rubbing

Sin30 = h / L

L = h / sin 30

WB = F d cos θ

WB = F L cos 30

WB = mf (h / sin30) cos 30

WB = mg h ctan 30

C) Ramp with rubbing

W sin 30 - fr = ma

N- Wcos30 = 0

W sin 30 - μ W cos 30 = ma

F = W (sin30 - μ cos30)

WC = mg (sin30 - μ cos30) h / sin30

Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

5 0

3 years ago

For an object to be classified as a ____, it must meet certain definite criteria: It must be massive enough to pull itself into

Orlov [11]

Answer:

a planet

Explanation:

a planet is one which exerts these properties and therefore is the answer

5 0

3 years ago