Because noble gases has full of 8 electron in her external cell......
Answer:
m = 0.0125 kg
Explanation:
Let us apply the formula for the speed of a wave on a string that is under tension:

where F = tension force
μ = mass per unit length
Mass per unit length is given as:
μ = m / l
where m = mass of the string
l = length of the string
This implies that:

Let us make mass, m, the subject of the formula:

From the question:
F = 20 N
l = 4.50 m
v = 85 m/s
Therefore:

Answer:
103.1 V
Explanation:
We are given that
Initial circumference=C=168 cm

Magnetic field,B=0.9 T
We have to find the magnitude of the emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.
Magnetic flux=
Circumference,C=

cm



When t=0



E=

t=8 s
B=0.9


Answer:
Final temperature, 
Explanation:
Given that,
Mass of silver ring, m = 4 g
Initial temperature, 
Heat released, Q = -18 J (as heat is released)
Specific heat capacity of silver, 
To find,
Final temperature
Solution,
The expression for the specific heat is given by :





So, the final temperature of silver is 21.85 degrees Celsius.
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