Initially, mg = kx. K = mg/x = 700/0.5x10^-3 = 1400000N/m. From second condition, applying work-energy theorem, potential enery- elastic potential energy = change in kinetic energy. Now change in kinetic energy is 0 since initial and final velocities are 0m/s. Therefore, potential energy = elastic potential energy. mgh = (1/2) * k* x^2. x^2 = 2(mg)h/k = 2 x 700 x 1.3/ 1400000. x = 0.036m. Hope it's clear.
Answer:
Explanation:
a) Using the equation of motion
S = ut + 1/2gt²
S is the distance of fall
g is the acceleration due to gravity
t is the time taken
Given S = 12.0m, g = 9.81m/s^2, un= 0m/s
12 = 0+1/2(9.81)t²
12 = 4.905t²²²
t² = 12/4.905
t² = 2.446
t = √2.446
t = 1.56secs
b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt
v = 0+9.81(1.56)
v = 15.34m/s
Hence the frog is falling at the rate of 15.34m/s
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Gs*rs^2 = gm*rm^2
<span>rm = rs*√gs/gm </span>
<span>rm = 6370*√9.83/(9.83-0.009) = 6372.92 </span>
<span>mountain observatory is placed at an altitude worth 2920 m asl
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Answer:
4,524,660 N
Explanation:
Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.
m/9 = (1026 kg/m³) (50 m³)
m = 461,700 kg
mg = 4,524,660 N