<span>...a concordant intrusion.
In geology, "concordant" means the same as "sill" -- or, an intrusion that has gotten in between older layers of rock (or even beds of volcanic lava). An intrusion with boundaries parallel to layering in surrounding rocks suggests this, meaning it is considered to be a concordant intrusion.</span>
Answer:
Kinda? Depends what the question is fully asking
Explanation:
Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.
So that -2 m/s thing after one second will be going -1 m/s.
After another second it'll be going 0 m/s.
After another itll be going +1 m/s and so on.
So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.
Explanation:
It is given that,
A particle starts from rest and has an acceleration function as :

(a) Since, 
v = velocity




(b) 
x = position



(c) Velocity function is given by :


t = 1 seconds
So, at t = 1 second the velocity of the particle is zero.
False
Balanced forces mean that there is no net force acting on the object. therefore, the object will not accelerate.
Answer:

Explanation:
Given that:
- mass of meteoroid,

- radial distance from the center of the planet,

- mass of the planet,

<u>For gravitational potential energy we have:</u>

substituting the respective values:

