Answer:
a)15 N
b)12.6 N
Explanation:
Given that
Weight of block (wt)= 21 N
μs = 0.80 and μk = 0.60
We know that
Maximum value of static friction given as
Frs = μs m g = μs .wt
by putting the values
Frs= 0.8 x 21 = 16.8 N
Value of kinetic friction
Frk= μk m g = μk .wt
By putting the values
Frk= 0.6 x 21 = 12.6 N
a)
When T = 15 N
Static friction Frs= 16.8 N
Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.
Friction force = 15 N
b)
When T= 35 N
Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N
Friction force = 12.6 N
Answer:
Explanation:
Position of charge 3q is x = 0
position of charge -2q is x = a
so here we know that
when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge
So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q
so here we can say
so we will have
so the x coordinate of this position is given as
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
-----------------------------------------------------------------
20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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Answer:
Found in the nucleus, Has mass of one amu
Given:
Horizontal distance between two boats = x = 14 m
One boat is at trough, the other is at crest.
As there is no crests between them meaning the boat are next to each other.
Wavelength is the distance between two consecutive crests/troughs = w
The distance between a crest and a trough next to it = w/2
Complete cycles = c = 5
Time taken for c cycles = t = 15 s
Vertical distance between two boats = y = 2.4 m
To find:
wavelength = w = 2x = 28 m
Amplitude = A = Displacement from mean to extreme position = y/2 = 1.2 m
Time period for one cycle = T = t/c = 15/5 = 3 s/cycle
frequency = 1/T = 1/3 = 0.33 hertz
speed = wavelength/Period = w/T = 28/3 = 9.33 m/s
