Answer:
(iv), (v), (vi) would be incorrect.
Explanation:
(iv) Force isn't transferred from one colliding object to another, but momentum can be.
(v) An object doesn't stop immediately a force stops acting on it. Think of a thrown ball.
(vi) For an object not to move, it means that the net force on the object is zero, and not necessarily that there are no forces acting on the object. For example, an object could be pushed on one side, and be pushed on the other side with an equal force in the opposite direction. The forces would cancel each other and the net force would be zero.
The rest should be correct.
A would be number 2. Newton's First Law states that an object at rest, will stay at rest and an object in motion, will stay in motion, unless acted upon by an unbalanced force. B would be number 3. His Second Law states that <span>the sum of the forces acting on a body is equal to the product of the mass of the body and the acceleration produced by the forces. And, C would be number 1. His Third Law states that for every action, there is an equal and opposite reaction. Hope this helps!</span>
Thank you for posting your question here at brainly. Below is Yoland's study:
<span>Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m.
</span>
I think the answer is "She can use constructive interference to generate a wave with an amplitude of 1.5 m."
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
Answer:
The correct answer is "0.32 mL".
Explanation:
The given values are:
Density of gold bar,
d = 19.3 g/mL
Mass of gold bar,
m = 6.3 grams
Now,
The volume will be:
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
