Answer:
0.8J
Explanation:
Given parameters:
Force = 20N
Compression = 0.08m
Unknown:
Spring constant = ?
Elastic potential energy = ?
Solution:
To solve this problem, we use the expression below:
F = k e
F is the force
k is the spring constant
e is the compression
20 = k x 0.08
k = 250N/m
Elastic potential energy;
EPE = 
k e² = x 250 x 0.08²
Elastic potential energy = 0.8J
Answer:
F = - K x
a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2
b) ω = (K/m)^1/2 angular frequency of SHM
ω = (133 / 1.3)^1/2 = 10.1 / sec
f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec
P = 1/f = .0157 sec
Answer:
2.6×10⁻³ N
Explanation:
From coulomb's law,
F = kq'q/r²................ Equation 1
Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.
From the question,
since q' = q
Then,
F = kq²/r²..................... Equation 2
Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,
Constant: k = 9×10⁹ Nm²/kg².
Substitute into equation 2
F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²
F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)
F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)
F = 2.6×10⁻³ N.
Answer:im just guessing d but i think its d though
Explanation:
it pretty obvious
