Planetary Nebulae hope that helps
The net force on the toy is 0 Newton.
The mass of the toy is 2 kg.
The force F₁ is the tension applied by one of the siblings on the toy is 8N.
Let us assume that the force applied by the other sibling is F₂, which is produced by the acceleration is 4m/s².
We know,
Force = Mass x acceleration
Putting values,
F₂ = 2 x 4
F₂ = 8N.
As we know that the forces are in opposite directions because the toy is being pulled by both the siblings in their respective directions which are opposite to each other.
Hence, the net force F is given by,
F = F₁ - F₂
F = 8-8
F 0N.
Hence, there is no net force on the toy.
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Answer:
independent” variable goes on the x-axis (the bottom, horizontal one) and the “dependent” variable goes on the y-axis (the left side, vertical one).
Explanation:
this is the answer I don't know if it's right
Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s
Answer:
M = ρ V = 9 gm/cm^ 3 * cm^3 = 27 gm
a = (V2 - V1) / t = (6 - 2) m/s / 12 s = 1/3 m/s^2 the acceleration
F = M a = 27 gm * 1/3 m/s^2 = 9 dynes net force applied