four particles connected by rigid rods of negligible mass where y1 = 5.70 m. the origin is at the center of the rectangle. The s
1 answer:
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Impulse on the object: -20 kg m/s
Explanation:
The impulse exerted on an object is a vector quantity equal to the product between the force exerted on the object and the time interval during which the force is applied.
The impulse exerted on an object is also equal to its change in momentum. Mathematically,
where:
m is the mass of the object
u is the initial velocity
v is the final velocity
For the object in this problem, we have
m = 5 kg
u = +6 m/s
v = +2 m/s
Therefore, its impulse is:
The negative sign means the direction of the impulse is opposite to the direction of motion of the object.
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Answer:
a
The x- and y-components of the total force exerted is
b
The magnitude of the force is
The direction of the force is
Clockwise from x-axis
Explanation:
From the question we are told that
The magnitude of the first charge is
The magnitude of the second charge is
The position of the second charge from the first one is
The magnitude of the third charge is
The position of the third charge from the first one is
The position of the third charge from the second one is
The force acting on the third charge due to the first and second charge is mathematically represented as
Substituting values
The magnitude of is mathematically evaluated as
The direction is obtained as
Answer:
The new voltage between the parallel plates of the capacitor is 18V, because for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage) between the plates.
Explanation:
ΔV = E*Δd
Where;
ΔV is the change in potential difference
Δd is the change in the distance between the parallel plates
E is the electric field potential.
Assuming a constant electric field;
when the spacing between the capacitor plates is doubled, d₂ = 2d₁
v₂ = (v₁*d₂)/(d₁)
v₂ = (v₁*2d₁)/(d₁)
v₂ = 2v₁
v₂ = 2(9) = 18 V
Therefore, for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage).