Answer:
0.231 m/s
Explanation:
m = mass attached to the spring = 0.405 kg
k = spring constant of spring = 26.3 N/m
x₀ = initial position = 3.31 cm = 0.0331 m
x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m
v₀ = initial speed = 0 m/s
v = final speed = ?
Using conservation of energy
Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy
(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²
m v₀² + k x₀² = m v² + k x²
(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²
v = 0.231 m/s
Answer:
Maximum Tension=224N
Minimum tension= 64N
Explanation:
Given
mass =8 kg
constant speed = 6m/s .
g=10m/s^2
Maximum Tension= [(mv^2/ r) + (mg)]
Minimum tension= [(mv^2/ r) - (mg)]
Then substitute the values,
Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N
Minimum tension= [8 × 6^2)/2 -(8×9.8)]
=64N
Hence, Minimum tension and maximum Tension are =64N and 2224N respectively
Answer:
a) t = 1.47 h b) t = 1.32 h
Explanation:
a) In this problem the plane and the wind are in the same North-South direction, whereby the vector sum is reduced to the scalar sum (ordinary). Let's calculate the total speed
v = 
f - 
v = 585 -32.1
v = 552.9 km / h
We use the speed ratio in uniform motion
v = x / t
t = x / v
t = 815 /552.9
t = 1.47 h
b) We repeat the calculation, but this time the wind is going in the direction of the plane
v= f -
v 585 + 32.1
v = 617.1 km / h
t = 815 /617.1
t = 1.32 h
