Answer:
6 m/s East
Explanation:
Momentum = Mass . Velocity
26100 = 4350. Velocity
Velocity = 6m/s East
The three R's<span> – reduce, reuse and recycle – all help to cut down on the amount of waste we throw away. They conserve natural </span>resources<span>, landfill space and energy.
Hope that helped :)</span>
Answer:
Most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.
Explanation:
- In a period as we advance from left to right, the number of valence electrons in the same shell increases due to which the effective nuclear charge increases and thus the atomic size decreases.
- In d-block atomic radius initially decreases then remains constant and increases towards the end.
- As one moves from Sc (scandium) to Zn (zinc), the effective nuclear charge increases by a factor of 1, this is because:
- The number of electrons are low in the inner shell.
- The shielding power of d-orbital is low.
- Inter electronic repulsions will be operating at a value less than the nuclear charge, which will result in decrease in atomic radii.
- As the number of electrons in the inner orbital increases the outer electrons repel one another which enables them to push away.
- Although d-orbital has less shielding power, the number of electrons present in it are high. Hence, the electron-electron repulsive force becomes dominant, this results in an increase in the atomic radii.
Therefore, most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.
Learn more about the periodic table here:
<u>brainly.com/question/9238898</u>
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F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
If we let
p as the directed multigraph that has no isolated vertices and has an Euler circuit
q as the graph that is weakly connected with the in-degree and out-degree of each vertex equal
The statement we have to prove is
p ←→q (for biconditional)
Since
p → q (assuming that p is strongly connected to q)
q ← p (since p is strongly connected to q)
Therefore, the bicondition is satisfied
