Answer:
F = - K X force constant for spring
a = F / m maximum acceleration
F = 4.5 kg * 26 m/s^2 = 117 Newtons
(A) K = 117 N / .038 m = 3079 N/m
ω = (K/M)^1/2 = (117/5)^1/2 = 4.84 / sec
(B) f = ω / 2 pi = 4.84 / 6.28 = .77 /sec
(C) P = 1 / f = 1/.77 = 1.30 sec
Answer:
oh for real?
Explanation:
The solubility of glucose at 30°C is
125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.
Answer:
88 m/s
Explanation:
To solve the problem, we can use the following SUVAT equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
d is the distance covered
For the car in this problem, we have
d = 484 m is the stopping distance
v = 0 is the final velocity
is the acceleration
Solving for u, we find the initial velocity:
Answer:
Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.
Explanation:
