Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Answer:
R=4Ω
Explanation:
R1 and R3 are parrallel
so we have : R=R1*R2/R1+R2
R=6*3/6+3
R=18/9
R=2Ω
R2 and 2Ω are in series,so we have
R=R2+2
R=2+2
R=4Ω
Yes it is concidered to be apart of the cars engine.
Answer: 10.3m/s
Explanation:
In theory and for a constant velocity the physics expression states that:
Eq(1): distance = velocity times time <=> d = v*t for v=constant.
If we solve Eq (1) for the velocity (v) we obtain:
Eq(2): velocity = distance divided by time <=> v = d/t
Substituting the known values for t=15s and d=155m we get:
v = 155 / 15 <=> v = 10.3
We can use constant acceleration equation to solve for the distance.
Formula is:
Vf^2 = Vi^2 + 2ad
where Vf^2 is final velocity squared, Vi^2 is initial velocity squared, a is acceleration (or deceleration) and d is the distance.
we want the car to come to complete stop, that is, Vf^2 be equal to zero.
Therefore, the equation becomes 0 = Vi^2 + 2ad. Solving for d we get:
d = (-(Vi)^2)/2a). We can ignore the minus sign since acceleration is really deceleration.
We know initial velocity (23m/s) and we know acceleration (max= 300 m/s^2). Plugging these in, we get:
d = ((23 m/s)^2)/(2* 300m/s^2) = <span>0.88m </span>
<span>hope that helps</span>
