A 50gk boy on a rough horizontal ground the coefficeint of static friction is 0.68
1 answer:
Given that,
Mass of a boy = 50 kg
The coefficient of static friction = 0.68
To find,
Let us assume we need to find the maximum static friction between the boy and the ground.
Solution,
The formula for the maximum static friction between the two objects is given by :
Where
N is normal force
Substitute all the values,
Therefore, the maximum static friction between the boy and the ground is 340 N.
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Answer:
minimum initial velocity is 21.35 m/s
Explanation:
given data
distance S = 30 m
height h = 30 m
maximum acceleration a = 2 m/s²
to find out
minimum initial velocity that your friend could have thrown the object to enable you to catch
solution
first we get here time with the help of second equation of motion
time = ..................1
put her value we get
time =
time = 5.477 second
and that is time which tossed object must be take so we apply here again second equation of motion that is
-S = ut - 0.5 × gt² .......................2
-30 = u× 5.477 - 0.5 ×9.8×5.477²
solve it we get
u = 21.35 m/s
so minimum initial velocity is 21.35 m/s
Answer:
n = 1810
A = 25 mm
Explanation:
Given:
Lateral force amplitude, F = 25 N
Frequency, f = 1 Hz
mass of the bridge, m = 2000 kg/m
Span, L = 144 m
Amplitude of the oscillation, A = 75 mm = 0.075 m
time, t = 6T
now,
Amplitude as a function of time is given as:
or amplitude for unforce oscillation
or
or
Now, provided in the question Amplitude of the driven oscillation
the value of the maximum amplitude is obtained
thus,
Now, for n people on the bridge
Fmax = nF
thus,
max amplitude
or
n = 1810
hence, there were 1810 people on the bridge
b)
since the effect of damping in the millenium bridge is 3 times
thus,
b=3b
therefore,
or
or
or
A = 0.025 m = 25 mm