Answer:
30 N
Explanation:
there are two forces act on the bar:
- weight of 1.5 kg mass, w = mg = 15 N
- weight of the bar, wb
for balance,
w * Lw = wb * Lwb
Lw = length of bar from the mass to the pivot
Lwb = lenght of bar from the center of the bar to the pivot
15 * 20 = wb * (50-20)
300 = wb * 30
wb = 300/30 = 30 N
Answer:
hello your question is incomplete below is the complete question
A 2.0 kg block starts from rest on a positive x axis 3.0m from the origin and thereafter has an acceleration given by a= 4.0i - 3.0jin m/s2. At the end of 2.0 s, its angular momentum about the origin is ______.
answer : L = ( 2 kg ) [(-18m^2/s)k]
Explanation:
velocity vector(v) = (4.0i - 3.0j )(2.0s)
position vector(r) = ( 3.0m)i
determine the angular momentum about the origin
L = m ( r * v )
attached below is the detailed solution
Answer:
10.2 m
Explanation:
The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:
where
y is the position of the m-th minimum
m is the order of the minimum
D is the distance of the screen from the slit
d is the width of the slit
is the wavelength of the light used
In this problem we have:
is the wavelength of the light
is the width of the slit
m = 13 is the order of the minimum
is the distance of the 13th dark fringe from the central maximum
Solving for D, we find the distance of the screen from the slit:
There is friction between the two.
Answer:
Explanation:
Given that,
We have two capacitors connected in series
C1=2.0-μF
C2=4.0-μF
Then the equivalent of their series connection
1/Ceq = ½ + ¼
1/Ceq= (2+1)/4
1/Ceq=¾
Taking the reciprocal
Ceq= 4/3 μF
The capacitors are connected to a battery of 1kv
V=1000Volts
We know that,
Q=CV
Where Q is charge
C is capacitance and
V is voltage
Then, Q=4/3 ×1000
Q=4000/3 -μC
Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF
After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,
For parallel connection, they have the same voltage but different charges.
When connected in parallel, there is a charge redistribution,
And the total charge will be 2•4000/3=8000/3 -μF
Then, Q1 +Q2= 8000/3 μF
Now the charge on each capacitor will be, let them have a common voltage V
Q=CV
Then, Q1=C1V
Q1= 2×V=2V
Q2= 4×V=4V
Then, Q1+Q2=8000/3
4V+2V=8000/3
6V=8000/3
V=8000/(3×6)
V=4000/9
V=444.44Volts
Now, Q1=2V
Q1=2×4000/9
Q1=8000/9 μF
Also, Q2=4V
Q2=4×4000/9
Q2=16000/9 μF