Answer: 12Mg/h
Explanation:
Let the spring is compressed by a distance x,before the lift stops,then
Mg(h+x)= 1/2 kx^2 ............... 1
Kx - Mg = M ( 5g ) ............ 2
Make x the subject in equation 2
Kx = 5Mg + Mg
Kx = 6Mg
x = 6Mg/k ............ 3
Put equation 3 into 1
Mg ( h + x ) = 1/2 kx^2
Mgh + Mgx = 1/2kx^2
Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2
Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2
h =18Mg/k - 6Mg/h
K = 12Mg/h
If the pressure above a solution containing a gas solute is reduced, the limit of the gas's solubility will decrease.
Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft
Answer:
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