Answer:
A) t = 29.3 s
B) V = 58.56 ft/s
C) a_net = 2.3 ft/s²
Explanation:
A) The formula for radial acceleration is given as;
a_c = V²/R
We are given;
Radius;R = 350 ft
So, a_c = V²/350
Where V is velocity
Tangential acceleration;a_t = 2 ft/s²
Formula for net acceleration is;
a_net = √((a_c)² + (a_t)²)
We are given a_net = 10 ft/s²
Thus;
10 = √(V²/350)² + (2²)
10² = V⁴/350² + 4
100 - 4 = V⁴/350²
96 × 350² = V⁴
V = 58.56 ft/s
Now, formula for angular velocity is;
ω = V/r
ω = 58.56/350
ω = 0.1673 rad/s
Angular acceleration is given by;
α = a_t/r
α = 2/350
α = 0.00571 rad/s²
Time needed will be gotten from the formula;
t = ω/α
t = 0.1673/0.00571
t = 29.3 s
B) we are told total acceleration is 10 ft/s², thus it's the same as velocity gotten earlier which is 58.56 ft/s
C) we are told that the speed is now 20 ft/s
Thus;
a_c = 20²/350
a_c = 1.1429 ft/s²
Since a_net = √((a_c)² + (a_t)²)
We are given a_t = 2 ft/s²
Thua;
a_net = √(1.1429² + 2²)
a_net = √5.30622041
a_net = 2.3 ft/s²
It is most like formed by convergent boundaries
Answer:
Diameter of the piston would be 0.71 m (71.1 cm)
Explanation:
From the principle of pressure;
= 
Let 
= 2903.57 lb, 
= 24.41 lbs, 
= 3.26 cm = 0.0326 m.
= 
= 
x 
= 0.00334 
So that:
= 
= 
= 0.3973
= 0.4
The radius of the piston can be determined by:
=
0.3973 = x 
= 
= 0.1264
r = 
= 0.3555
r = 0.36 m
Diameter of the piston = 2 x r
= 2 x 0.3555
= 0.711
Diameter of the piston would be 0.71 m (71.1 cm).
2.12m/s is water and if it goes 1.5 seconds fast
