a. The force applied would be equal to the frictional
force.
F = us Fn
where, F = applied force = 35 N, us = coeff of static
friction, Fn = normal force = weight
35 N = us * (6 kg * 9.81 m/s^2)
us = 0.595
b. The force applied would now be the sum of the
frictional force and force due to acceleration
F = uk Fn + m a
where, uk = coeff of kinetic friction
35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)
uk = 0.533
Answer:
AB = 29
Explanation:
For a better understanding, we must work this problem in a graphic way. In the attached image we can see the solution.
First, we draw a vertical dotted lines from the point J & K to the line ML, then we can see two new portions with the same length. Then with this simple analysis:
2x = 39 - 19
x = 10
Then we know that x = 10, another important data to find the answer is that the AB line is located in the midpoints of the legs. We also can see the right triangle MJ and the dotted line.
Now for every single right triangle, no matter its size and relationship between the vertical and the horizontal lengths, if some point is located in the hypotenuse (leg) at the middle of its length. This will be proportional to the vertical and the horizontal cathetus, therefore we will have the middle point on those two lines.
So, the AB line will be the sum of JK plus two times 5
AB = 19 + 5 + 5 = 29
Explanation:
(a) From E=
r
2
k
e
Q
Q=
k
e
Er
2
=
(8.99×10
9
N⋅m
2
/C
2
)
(8.90×10
2
N/C)(0.750m)
2
=5.57×10
−8
C
But Q is negative since
E
→
points inward, so
Q=−5.57×10
−8
C=−55.7nC
(b) The negatve charge has a spherically symmetric charge distribution, concentric with the spherical shell
Answer:

Explanation:
first ball rolls on the porch by total distance

Then again it will move on horizontal floor

also in vertical direction it will drop down

so we have



so magnitude of net displacement of the ball is given as

