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
Answer:
B. normal force
Explanation:
Because there is no frictional or resistance force. However gravitational force is applied downroad from the center of the cup thus the contact force that is perpendicular to the surface that an object contacts which is the normal force exerted upward from the table that prevents an object from falling.
Answer:
I Will say the Answer is A
Explanation:
Given: Velocity of light c = 3.00 x 10⁸ m/s
Frequency f = 7.65 x 10⁷/s
Required: Wavelength λ = ?
Formula: λ = c/f
λ = 3.00 x 10⁸ m/s/7.65 x 10⁷/s
λ = 3.92 m
Hi there!
We can begin by solving for the linear acceleration as we are given sufficient values to do so.
We can use the following equation:
vf = vi + at
Plug in given values:
4 = 9.7 + 4.4a
Solve for a:
a = -1.295 m/s²
We can use the following equation to convert from linear to angular acceleration:
a = αr
a/r = α
Thus:
-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.
Now, we can find the angular displacement using the following:
θ = ωit + 1/2αt²
We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:
v = ωr
v/r = ω
9.7/0.61 = 15.9 rad/sec
Plug into the equation:
θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad
