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
I think it is c density and temperature
<u>We are given:</u>
Mass of the Steelhead(m) = 9 kg
Velocity of the Steelhead(v) = 16 m/s
<u>Calculating the Kinetic Energy:</u>
KE = 1/2mv²
replacing the variables
KE = 1/2 * 9 * (16)²
KE = 1152 Joules
Answer:
m = 0.164 kg
Explanation:
T (period)
k (force/spring constant)
m (mass)
T = 2*Pi*sqrt(m/k)
T/(2*Pi) = sqrt(m)/sqrt(k)
(T/(2*Pi))*sqrt(k) = sqrt(m)
m = ((T/(2*Pi))*sqrt(k))^2
m = 4.5*((1.2/(2*Pi)))^2
m = 0.1641403175
Answer:
To share a positive experience she had with a pen pal
Explanation:
Just read the story :)
