Answer:
The average drag force is 1.206 (-i) N
Explanation:
You have to apply the equations of<em> Impulse</em>:
I=FmedΔt
Where I and Fmed (the average force) are vectors.
The Impulse can also be expressed as the change in the <em>quantity of motion</em> (vector P)
I=P2-P1
P=mV (m is the mass and v is the velocity)
You can calculate the quantity of motion at the beggining and at the end of the given time:
Replace the mass in kg, dividing the mass by 1000 to convert it from g to kg.
P1=(0.179kg)(30.252m/s) i= 5.414 i kg.m/s
P2=0.179kg)(28.452m/s) i = 5.092 i kg. m/s
Where i is the unit vector in the x-direction.
Therefore:
I= 5.092 i - 5.414 i = -0.322 i
The average drag force is:
Fmed= I/Δt = -0.322 i/ 0.267s = -1.206 i N
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm
Answer:
v = 10 m/s
Explanation:
given,
Mass of large fish, M = 5 Kg
speed of swimming, v' = 1 m/s
mass of small fish, m = 500 g = 0.5 Kg
speed of the fish = v
using conservation of momentum
m v + M v' = M u' + m u
final speed of both the speed is zero.
- 0.5 x v + 5 x 1 = 0
negative sign is used because small fish is moving in opposite direction.
now,
0.5 v = 5
v = 10 m/s
hence, the speed of the small fish is equal to 10 m/s.
