The coefficient of friction must be 0.196
Explanation:
For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:
where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:
is the coefficient of friction between the tires and the road
m is the mass of the car
is the acceleration of gravity
v is the speed of the car
r is the radius of the curve
In this problem,
r = 750 m is the radius
is the speed
And solving for , we find the coefficient of friction required to keep the car in circular motion:
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The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.
Given:
Potential difference, V = 1.2 V
Charge on an electron, e = 1.6 × 10⁻¹⁹ C
Calculation:
We know that the work done to transport an electron from the positive to the negative terminal is given as:
W.D = (Charge on electron)×(Potential difference)
= e × V
= (1.6 × 10⁻¹⁹ C)×(1.2 V)
= 1.92 × 10⁻¹⁹ J
Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.
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Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,
<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,
Substituting,
Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,
Where here 
is the velocity after the collision.
Answer:
Q = 12.466μC
Explanation:
For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:
Solving for Q:
Taking special care of all units, we can calculate the value of the charge:
Q = 12.466μC
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