<h2>Circular path i.e 1</h2>
Explanation:
If the driver steer into circular path , the acceleration of car will be
acceleration = velocity²/radius of circle
Because the velocity and radius of circle are both constant at all points . Thus the acceleration will also remain constant .
In case 3 and 4 , the radius of path changes . Because velocity is constant . Therefore the acceleration will not remain constant
Observe that the given vector field is a gradient field:
Let 
, so that
Integrating the first equation with respect to 
, we get
Differentiating this with respect to 
gives
Now differentiating 
with respect to 
gives
Putting everything together, we find a scalar potential function whose gradient is 
,
It follows that the curl of is 0 (i.e. the zero vector).
Newtons law states F=ma
so F/m=a
in addition it makes logical sense the lighter an object the easier it is to accelerate its velocity form 0 to some quantity
Answer:
i think the answer would be:Jonas’ brother gets out of the cab of the truck and sits in the back of the truck with the furniture. With less mass, they should be able to push the truck to the gas station.
Explanation:
Answer:
Ball hit the tall building 50 m away below 10.20 m its original level
Explanation:
Horizontal speed = 20 cos40 = 15.32 m/s
Horizontal displacement = 50 m
Horizontal acceleration = 0 m/s²
Substituting in s = ut + 0.5at²
50 = 15.32 t + 0.5 x 0 x t²
t = 3.26 s
Now we need to find how much vertical distance ball travels in 3.26 s.
Initial vertical speed = 20 sin40 = 12.86 m/s
Time = 3.26 s
Vertical acceleration = -9.81 m/s²
Substituting in s = ut + 0.5at²
s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²
s = -10.20 m
So ball hit the tall building 50 m away below 10.20 m its original level
