Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
What is he minumum coating of thickness needed to ensure that lifght of waveelntght 5660 mbnd si
Answer:
Follows are the solution to the given question:
Explanation:
In this question the missing file of the circuit is not be which is defined in attached file please find it.
In Option 1, this statement is true because the current is on 
, that is the same.
In option 2, this statement is false because 
therefore it implies that Rcd is always larger then 
.
In option 3, this statement is true because the voltage of 
is always equal.
In option 4, this statement is true because 
is always smaller then 1 therefore,
is always equal to R1.
