Answer:
2.64 m/s
Explanation:
Given that a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna
Momentum = MV
Momentum of the yellow shark before collision = 600 × 3 = 1800 kgm/s
Momentum of the tun final before collision = 100 × 0.5 = 50 kgm/s
Total momentum before collision = 1800 + 50 = 1850 kgm/s
Let's assume that they move together after collision. Then,
1850 = ( 600 + 100 ) V
1850 = 700V
V = 1850 / 700
V = 2.64285 m/s
Therefore, the momentum of the shark after collision is 2.64 m/ s approximately
<h2>
Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>
Explanation:
The half-life 
of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.
In this case, we are given the half life of two elements:
beryllium-13: 
beryllium-15: 
As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?
We can find it out by the following expression:
Where 
is the amount we want to find:
Finally:
Therefore:
The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.
C₂H₃O₂⁻ is an anion.
<u>Explanation:</u>
NaC₂H₃O₂(s) → Na⁺(aq) + C₂H₃O₂⁻(aq)
NaC₂H₃O₂ when dissociated, yields Na⁺ and C₂H₃O₂⁻.
Anion is a negatively charged ion.
In this case, C₂H₃O₂⁻ is an anion.
Explanation:
Given that,
Mass = 0.254 kg
Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]
Force = 0.5 N
y = 0.628
We need to calculate the A and d
Using formula of A and d
.....(I)
....(II)
Put the value of 
in equation (I) and (II)
From equation (II)
Put the value of 
in equation (I) and (II)
From equation (II)
Put the value of 
in equation (I) and (II)
From equation (II)
Put the value of 
in equation (I) and (II)
From equation (II)
Hence, This is the required solution.
