Answer:
The phase constant is 7.25 degree
Explanation:
given data
mass = 265 g
frequency = 3.40 Hz
time t = 0 s
x = 6.20 cm
vx = - 35.0 cm/s
solution
as phase constant is express as
y = A cosФ ..............1
here A is amplitude that is = = = 6.25 cm
put value in equation 1
6.20 = 6.25 cosФ
cosФ = 0.992
Ф = 7.25 degree
so the phase constant is 7.25 degree
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Chicken wings, Chicken wings
Hotdog and baloney
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Explanation:
im sorry i had to do it now pls give me brainliest thank u and have a blessed day:)
Answer:
Incomplete question: "Each block has a mass of 0.2 kg"
The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s
Explanation:
Given data:
θ = angle of the surface = 37°
m = mass of each block = 0.2 kg
v = speed = 0.35 m/s
t = time to collision = 0.5 s
Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?
Change in momentum:
It is neccesary calculate the force:
Here, g = gravity = 9.8 m/s²
Answer:
Modeling tool or Align tool. it depends what type of sandbox platform you use
Explanation:
1
Answer:
f(t) = 28,7 [N]
Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.
The net force on +q₂ is the sum of the force of +q₁ on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)
The two forces have the same direction to the right of charge q₂, we have to add them
Then
f(t) = f₁₂ + f₃₂
f₁₂ = K * ( q₁*q₂ ) / (0,1)²
q₁ = + 8 μC then q₁ = 8*10⁻⁶ C
q₂ = + 3,5 μC then q₂ = 3,5 *10⁻⁶ C
K = 9*10⁹ [ N*m² /C²]
f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻² [ N*m² /C²]* C*C/m²
f₁₂ = 252*10⁻¹ [N]
f₁₂ = 25,2 [N]
f₃₂ = 9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²
f₃₂ = 78,75*10⁻³/ 2,25*10⁻²
f₃₂ = 35 *10⁻¹
f₃₂ = 3,5 [N]
f(t) = 28,7 [N]