<span>Before answering the questions I will list the three main laws of motion. The first law of motion states that an object at rest tends to stay at rest, and an object in motion tends to stay in motion in a straight line at a constant speed unless another force acts upon it. The second law of motion states that how fast an object picks up speed or slows down depends on the amount of force used and on its mass. The third law of motion states that for every action there is an equal and opposite reaction.
#1) When a propagule falls into the water, it displaces some water, which causes a small splash.
Answer: For this case it would be C) The Third Law of Motion.
#2) A heavier propagule will fall faster and hit the water with more force, causing a bigger splash.
Answer: For this case it would be B) The Second Law of Motion.
#3) The drifting propagule tends to stay in motion. It will only get stuck in the mud and start to take root when it is really time to germinate.
Answer: For this case it would be C) The Third Law of Motion.
#4) When a propagule is ready to fall off of the parent tree, it must overcome the inertia that keeps it in place.
Answer: For this case it would be A) The First Law of Motion.
#5) Occasionally, another branch or leaf will rub against a seedling due to wind or water action, and cause it to fall prematurely.
Answer: For this case it would be C) The Third Law of Motion.
#6) A seedling falling into the water does so with more force than a leaf, since it has more mass.
Answer: For this case it would be B) The Second Law of Motion.
#7) As a seedling's roots begin to grow into the mud, they displace the soil particles, which are moved aside to make room for the new tree.
Answer: For this case it would be D) Friction.
#8) Manatees are large marine mammals that live in the mangroves. To scratch their itchy backs, they like to rub against the roots of the mangroves.
Answer: For this case it would be D) Friction.
I hope it helps, Regards. <span>
</span></span>
Explanation:
The question is sound. The frequency and wavelength are given so you can solve for wave's velocity. It's just a matter of changing the notation.
![v = \nu \lambda = (15×10^9\:\text{Hz})(0.2\:\text{m})](https://tex.z-dn.net/?f=v%20%3D%20%5Cnu%20%5Clambda%20%3D%20%2815%C3%9710%5E9%5C%3A%5Ctext%7BHz%7D%29%280.2%5C%3A%5Ctext%7Bm%7D%29)
![= 3×10^9\:\text{m/s}](https://tex.z-dn.net/?f=%3D%203%C3%9710%5E9%5C%3A%5Ctext%7Bm%2Fs%7D)
Answer:
Sodium sulfate(Na2SO4) is a salt
Explanation:
It is a neutralization reaction in which sulphuric acid (H2SO4) which is composed of sulfur, oxygen, and hydrogen. Sulphuric acid is strong acid reacts with 2 moles of Sodium hydroxide (in neutralization reaction 1 mole of sulphuric acid is equal to 2 moles of Sodium hydroxide) gives us 2 moles of water and sodium sulfate (Na2SO4) which is composed of sodium, sulfur, and oxygen.
Sodium sulfate (Na2SO4) is salt, when H2SO4 reacts with sodium hydroxide (due to hydrate) it will give us salt (Na2SO4)
Answer:
The total distance traveled is 736 m
Solution:
According to the question:
Initial velocity, v = 0
(since, the car is starting from rest)
![acceleration, a = 4 m/s^{2}](https://tex.z-dn.net/?f=acceleration%2C%20a%20%3D%204%20m%2Fs%5E%7B2%7D)
Time taken, t = 8 s
Now, the distance covered by it in 8 s is given by the second eqn of motion:
![d = vt + \farc{1}{2}at^{2}](https://tex.z-dn.net/?f=d%20%3D%20vt%20%2B%20%5Cfarc%7B1%7D%7B2%7Dat%5E%7B2%7D)
![d = 0.t + \farc{1}{2}4\times 8^{2} = 128 m](https://tex.z-dn.net/?f=d%20%3D%200.t%20%2B%20%5Cfarc%7B1%7D%7B2%7D4%5Ctimes%208%5E%7B2%7D%20%3D%20128%20m)
Now, to calculate the velocity, we use eqn 1 of motion:
v' = v + at
v' = 0 + 4(8) = 32 m/s
Now, the distance traveled by the car with uniform velocity of 32 m/s for t' = 19 s:
distance, d' = v't'
![d' = 32\times 19 = 608 m](https://tex.z-dn.net/?f=d%27%20%3D%2032%5Ctimes%2019%20%3D%20608%20m)
Total distance traveled = d + d' = 128 + 608 = 736 m
Answer:
a particular kind of matter with uniform properties.
Explanation: