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ivanzaharov [21]
3 years ago
12

The electrons between atoms in metallic bonds

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

Answer:

In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize.

emmasim [6.3K]3 years ago
4 0

Answer:

<em>b) allow for bonding metals to be stable.</em>

Explanation:

got it right on a quiz on edg.

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goldenfox [79]

Answer:

its B

Explanation:

the movement of charged particles in a conductor

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A scientific law seeks to explain why an event occurred. True or false
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The answer would be A) True.
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Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person we
jeka57 [31]

(a) 4.14 rad/s^2

The relationship beween centripetal acceleration and angular speed is

a=\omega^2 r

where

\omega is the angular speed

r is the radius of the circular path

Here we gave

a = 9g = 88.2 m/s^2 is the centripetal acceleration

r = 5.15 m is the radius

Solving for \omega, we find:

\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

v=\omega r

where

v is the linear speed

\omega is the angular speed

r is the radius of the circular path

In this problem we have

\omega=4.14 rad/s

r = 5.15 m

Solving the equation for v, we find

v=(4.14 rad/s)(5.15 m)=21.3 m/s

7 0
3 years ago
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How would you get your friend to change his/her mind from using PED's?​
kvasek [131]

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Watch funny videos and start eating a lot of candy

6 0
3 years ago
If an athlete leaps vertically at 4.0m/s, what maximum height does he reach?
joja [24]
Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: g=-9.81~m/s^2, and where the negative sign means it points downwards, against the direction of the motion.

Therefore, we can use the following formula to solve the problem:
v_f^2 = v_i^2 + 2gS
where v_i=4~m/s is the initial vertical velocity of the athlete, v_f=0 is the vertical velocity of the athlete at the maximum height (and v_f=0~m/s at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
S= \frac{v_f^2-v_i^2}{2g}=0.82~m

3 0
3 years ago
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