“Don't hand that holier than thou line to me” is what the asymptote
said to the removable discontinuity.
The distance between the
curve and the line where it approaches zero as they tend to infinity is the line in the asymptote
of a curve. This is unusual for modern authors but in some
sources the requirement that the curve may not cross the line infinitely often
is included.
The point that does not fit the rest of the graph or is
undefined is called a removable discontinuity. By filling in a single
point, the removable discontinuity can be made connected.
Answer: The trip takes 
Explanation:
Velocity 
is the variation of the position of a body (distance traveled 
) with time 
:
In this case, the car travels a distance 
at a velocity 
and we need to find the time it takes the trip.
Isolating :
Finally:
An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum and kinetic energy are both conserved.
<h3>Explain about the Elastic Collision?</h3>
A collision between two bodies in physics is referred to as an elastic collision if their combined kinetic energy stays constant. There is no net conversion of kinetic energy into other forms, such as heat, noise, or potential energy, in an ideal, fully elastic collision
An example of an elastic collision is when two balls collide at a pool table. It is an elastic collision when you throw a ball on the ground and it bounces back into your hand because there is no net change in the kinetic energy.
If there is no kinetic energy lost in the impact, the collision is said to be perfectly elastic. A collision is considered to be inelastic if any of the kinetic energy is converted to another kind of energy during the collision.
To learn more about Elastic Collision refer to:
brainly.com/question/7694106
#SPJ4
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at 
Solving for the angle, we get
or
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