In a lab, a student drags a shoe across the floor at constant speed. If the coefficient of static friction between the floor and
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Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Explanation:
Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.
Now, according to the law of conservation of energy the formula used is as follows.
As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.
Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Answer:
The charge q₃ must be placed at X = +2.5 cm
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1µC= 10⁻6 C
1cm= 10⁻² m
Data
k = 8.99*10⁹ N×m²/C²
q₁ =+3 µC =3*10⁻⁶ C
q₂ = -10 µC =-10*10⁻⁶ C
q₃= +6µC =+6*10⁻⁶ C
d₁ = 3cm =3×10⁻² m
d₂ = 4cm = 4×10⁻² m
Graphic attached
The attached graph shows the field due to the charges:
E₁:Field at point P due to charge q₁. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).
E₂: Field at point P due to charge q₂. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).
Problem development
E₃: Field at point P due to charge q₃. As the charge q₃ is positive, the field leaves the charge.
The direction of E₃ must be (- x) so that the electric field can be equal to zero at point P since E₁ and E₂ are positive, then, q₃must be located to the right of point P.
We make the algebraic sum of fields at point P due to the charges q1, q2, and q3:
E₁+E₂-E₃=0
We eliminate k
We replace data
we eliminate 10⁻⁶
The charge q₃ must be placed at X = +2.5 cm
