A. Impulse is simply the product of Force and time.
Therefore,
I = F * t --->
1
where I is impulse, F is force, t is time
However another formula for solving impulse is:
I = m vf – m vi --->
2
where m is mass, vf is final velocity and vi is initial
velocity
Therefore using equation 2 to solve for impulse I:
I = 2000kg (0) – 2000kg (77 m/s)
I = -154,000 kg m/s
B. By conservation of momentum, we also know that Impulse
is conserved. That means that increasing the time by a factor of 3 would still
result in an impuse of -154,000 kg m/s. So,
I = F’ * (3 t) = -154,000 kg m/s
Since t is multiplied by 3, therefore this only means
that Force is decreased by a factor of 3 to keep the impulse constant,
therefore:
(F/3) (3t) = -154,000 kg m/s
Summary of Answers:
A. I = -154,000 kg m/s
B. Force is decreased by factor of 3
Sonar, originally an <span>acronym for SOund Navigation And Ranging
</span>a technique that uses sound<span> propagation.</span>
B. The suns gravity
hope this helps
Answer:
60 days.
Explanation:
Let the original mass (N₀) = 1 g
Amount remaining (N) = 6.25% of its original mass
= 6.25% × 1
= 6.25/100 × 1
= 0.0625 g
Half life (t½) = 15 days
Time (t) =?
Next, we shall determine the rate of decay. This can be obtained as follow:
Decay constant (K) = 0.693/ half life
K = 0.693 / t½
Half life (t½) = 15 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 15
K = 0.0462 / day
Finally, we shall determine the time taken for the sample of the isotope to decay to 6.25% of its original mass.
This can be obtained as follow:
Original amount (N₀) = 1 g
Amount remaining (N) = 0.0625 g
Decay constant (K) = 0.0462 / day
Time (t) =?
Log (N₀/N) = Kt/2.3
Log (1/0.0625) = 0.0462 × t / 2.3
Log 16 = 0.0462 × t / 2.3
1.2041 = 0.0462 × t /2.3
Cross multiply
0.0462 × t = 1.2041 × 2.3
Divide both side by 0.0462
t = (1.2041 × 2.3)/0.0462
t = 59.9 ≈ 60 days
Therefore, the time taken for the sample of the isotope to decay to 6.25% of its original mass is 60 days
C. The cell wall supports the plant cell as it contains celluose which is a very strong substance and even thought chloroplast is in both that is the only available answer. Hope it helps
