There are no true statements on that list.
did you tried first if you did I can help
Velocity = fλ
where f is frequency in Hz, and λ is wavelength in meters.
<span>2.04 * 10⁸ m/s = 5.09 * 10¹⁴ Hz * λ </span>
<span>(2.04 * 10⁸ m/s) / (5.09 * 10¹⁴ Hz ) = λ </span>
<span>4.007*10⁻⁷ m = λ </span>
<span>The wavelength of the yellow light = 4.007*10⁻⁷ m<span> </span></span>
Answer:
t = (ti)ln(Ai/At)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Explanation:
Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t
At = Ai/2^n ..... 1
Where n is the number of half-life that have passed.
n = t/half-life
Half life = 14
n = t/14
At = Ai/2^(t/14)
From equation 1.
2^n = Ai/At
Taking the natural logarithm of both sides;
nln(2) = ln(Ai/At)
n = ln(Ai/At)/ln(2)
Since n = t/14
t/14 = ln(Ai/At)/ln(2)
t = 14ln(Ai/At)/ln(2)
Ai = 800
At = 50
t = 14ln(800/50)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Let half life = ti
t = (ti)ln(Ai/At)/ln(2)
IF the bag is suspended freely in mid-air and free to move
without friction, then the linear momentum is the same both
before and after the hit.
Momentum of the bullet before the hit = ( m₁ v₁ ).
Momentum of the (bullet + bag) after the hit = (m₁ + m₂) (v₂) .
Momentum is conserved, so we can say that ...
( m₁ v₁ ) = (m₁ + m₂) (v₂)
v₂ = ( m₁ v₁ ) / (m₁ + m₂)
Usually the mass of the bag is much much more than the mass
of the bullet, so the mass of the bullet can be ignored after the
hit, and the formula is very closely ...
v₂ = v₁ ( m₁ / m₂ ) .
Stop apologizing for your English. It's better than many native
Americans writing on this site. Be proud.
