Metallic bonds are responsible for many properties of metals, such as conductivity. This is because the bonds can shift because valence electrons are held loosely and move freely. That is option C.
<h3>What are metallic bonds?</h3>
Metallic bonds are defined as those bonds that causes the electrostatic attraction between metal cations and delocalized electrons of another metallic substance.
The characteristics of a metallic compound with metallic bonds include the following:
- thermal and electrical conductivity,
The metallic bonds of these metallic atoms gives them conductivity features because the electrons from the outer shells of the metal atoms are delocalised , and are free to move through the whole structure.
Learn more about metals here:
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Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Answer:
(c) 16 m/s²
Explanation:
The position is ![r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}](https://tex.z-dn.net/?f=r%28t%29%20%3D%20%5B3.0%20%5Ctext%7B%20m%7D%20-%20%284.00%20%5Ctext%7B%20m%2Fs%7D%29t%5D%5Chat%7Bi%7D%20%2B%20%5B6.0%20%5Ctext%7Bm%7D%20-%20%288.00%20%5Ctext%7B%20m%2Fs%7D%5E2%20%29t%5E2%20%5D%5Chat%7Bj%7D)
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The velocity is the first time-derivative of <em>r(t).</em>
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The acceleration is the first time-derivative of the velocity.
Since <em>a(t)</em> does not have the variable <em>t</em>, it is constant. Hence, at any time,
Its magnitude is 16 m/s².
Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;
Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.
