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:
Answer:
Using the given values
F = K q^2 / r^2 = 9 * 10E9 * (1.6 * E-19)^2 / (5.18 * E-15)^2 N
E = 9 * 1.6^2 / 5.18^2 * 10 = 8.5 N
•THAT THE PROPAGATION OF SOUND WAVES NEED MEDIUM TO TRAVEL
•THE MEDIUM SHOULD POSSES ELASTICITY
•FOR THE FASTER PROPAGATION OF SOUND THE PARTICLES SHOULD BE VERY CLOSE TO EACH OTHER
Answer:
a) = 258352.5J
b) = 23.63 m/s
c) = 1.8m
Explanation:
Data;
Mass = 925kg
Distance (s) = 28.5m
Force constant (k) = 8.0*10⁴ N/m
g = 9.8 m/s²
a) = work = force * distance
But force = mass * acceleration
Force = 925 * 9.8 = 9065N
Work = F * s = 9065 * 28.5 = 258352.5J
b) acceleration (a) = (v² - u²) / 2s
a = v² / 2s
v² = a * 2s
v² = 9.8 * (2 * 28.5)
v² = 9.8 * 57
v² = 558.6
v = √(558.6)
V = 23.63 m/s
C). The work stops when the work done to raise the spring equals the work done to stop it by the spring
W = ½kx²
258352.5 = ½ * 8.0*10⁴ * x²
(2 * 258352.5) = 8.0*10⁴x²
516705 = 8.0*10⁴x²
X² = 516705 / 8.0*10⁴
X² = 6.46
X = √(6.46)
X = 2.54m
The compression was about 2.54m
Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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