Hydrogen gas (H2) can be found in trace amounts in Earth’s atmosphere. Which of these statements describes a physical property o
1 answer:
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Answer:
The correct answer is a
Explanation:
At projectile launch speeds are
X axis vₓ = v₀ = cte
Y axis = v_{oy} –gt
The moment is defined as
p = mv
For the x axis
pₓ = mvₓ = m v₀ₓ
As the speed is constant the moment is constant
For the y axis
p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)
Speed changes over time, so the moment also changes over time
Let's examine the answer
i True
ii False. The moment changes with time
The correct answer is a
Answer:
(A) 0.279N at angle 38.02°
(B) 0.701N
(C) 14.19°
Explanation:
(A) The net force on q3 is given as:
F = Fxi + Fyj
Fx is the x component of the force
Fy is the y component of the force
Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0
Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx
First let us find y and angle x from the diagram.
Using Pythagoras theorem,
y² = 0.3² + 0.4²
y² = 0.25
y = 0.5m
Using SOHCAHTOA to find x,
sinx = 0.4/0.5
x = 53.13°
Electrostatic force, F is given as:
F = kqQ/r²
Where k = Coulumbs constant
F(1,3) = (k*q1*q3) / r²
F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)
F(1, 3) = 0.288N
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = 0.45N
Therefore,
Fx = -0.288cos36.87 + 0.45
Fx = 0.22N
Fy = 0.288cos53.13
Fy = 0.172N
=> F = 0.22i + 0.172j
The magnitude of the force will be
F(mag) = √(0.22² + 0.172²)
F(mag) = 0.279N
The direction of the force makes will be
tanθ = Fy/Fx
tanθ = 0.172/0.22 = 0.781
θ = 38.02° to the x axis.
(B) q2 = - 2.0 * 10^(-6)
This implies that:
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = -0.45N
Therefore,
Fx = -0.288cos36.87 - 0.45
Fx = -0.68N
Fy = 0.172N
=> F = - 0.68i + 0.172j
The magnitude of the force will be
F(mag) = √((-0.68)² + 0.172²)
F(mag) = 0.701N
(C) The direction of the force makes will be
tanθ = 0.172/0.68
θ = 14.19° to the x axis
