What fraction of stipend triangle is a shaded triangle? What fraction of the spotted triangle is a shaded triangle? Use >,
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
The answer is: [C]: "elasticity" .
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Answer:
No work is performed or required in moving the positive charge from point A to point B.
Explanation:
Lets take
Q= Positive charge which move from point A to point B along
Voltage difference,ΔV =V₁ - V₂
The work done
W = Q . ΔV
Given that charge is moved from point A to point B along an equipotential surface.It means that voltage difference is zero.
ΔV = 0
So
W = Q . ΔV
W = Q x 0
W= 0 J
So work is zero.
Answer:
The force is the same
Explanation:
The force per meter exerted between two wires carrying a current is given by the formula
where
is the vacuum permeability
is the current in the 1st wire
is the current in the 2nd wire
r is the separation between the wires
In this problem
Substituting, we find the force per unit length on the two wires:
However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.
The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).
