A bus travels 300 km south along a straight path with an average velocity of 85 km/h to the south. The bus stops for 22 minutes.
1 answer:
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Answer:
a) 1.6*10^6 V
b) 13.35*10^6 V
Explanation:
The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:
(1)
q1 = 3.90µC = 3.90*10^-6 C
q2 = -2.4µC = -2.4*10^-6 C
r1 = 1.25 cm = 0.0125 m
r2 = -1.80 cm = -0.018 m
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
You replace all the parameters in the equation (1):
hence, the total electric potential is approximately 1.6*10^6 V
b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:
r1 = 0.0150m - 0.0125m = 0.0025m
r2= 0.015m + 0.018m = 0.033m
Then, you replace in the equation (1):
hence, for y = 1.50cm you obtain V = 13.35*10^6 V
1) The coefficient of static friction is 0.980
2) The coefficient of kinetic friction is 0.908
Explanation:
1)
In the first situation, the box is still at rest. There are two forces acting on the box:
- The force of push, F, forward
- The force of static friction,
Since the box is in equilibrium we have
(1)
The value of frictional force changes from zero to a maximum value which is given by:
(2)
where
is the coefficient of static friction
m is the mass of the box
g is the acceleration of gravity
So, if the force F needed to put the box in motion is 48 N, it means that this is also the maximum value of the force of friction. So, we can combine eq.(1) and (2) to find the coefficient of static friction:
where:
F = 48 N
m = 5.0 kg
Substituting,
2)
In this second situation, the object is already in motion, so the equation of motion is:
where
F = 48 N is the force applied forward
is the acceleration of the box
is the force of kinetic friction, where
is the coefficient of kinetic friction
We can therefore rearrange the equation to find the coefficient:
Learn more about friction:
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