The actual answer is B) Chlorine
According to the Bohr Model diagram, the atom has seventeen electrons. This makes it Chlorine.
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The answer is C.
The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
Answer:
f=-1380N
Explanation:
A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and as it strikes the board, his hand has a velocity of 23.3 m/s. The master contacts the board for 0.0050 seconds
.the concluding part to the question should be
What is the impact force (impulse) on the board?
solution
from the Newton's second law of motion which states that
the rate of change in momentum is directly proportional to the force applied
f=m(v-u)/t
f=0.3(0-23.3)/0.005
f=-1380N
f=force impact
m=mass of the karates master's hand
t=time for the impact
v=0m/s final velocity
u=initial velocity
Regardless of what direction an object is moving, the acceleration
due to gravity is always directed toward the center of the Earth.
That's the direction commonly known as "down".
The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
The given parameters;
- <em>initial temperature of metals, = </em>
<em /> - <em>initial temperature of water, = </em>
<em> </em> - <em>specific heat capacity of copper, </em>
<em> = 0.385 J/g.K</em> - <em>specific heat capacity of aluminum, </em>
= 0.9 J/g.K - <em>both metals have equal mass = m</em>
The quantity of heat transferred by each metal is calculated as follows;
Q = mcΔt
<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
Learn more here:brainly.com/question/15345295