Answer:
20 cm to the right of the center or 20+50 = 70 cm from the left side.
Explanation:
The length of meter stick is 1 m = 100 cm
Balance point on 50 cm
From the center the 20 N weight is 50-20 = 30 cm
Torque is obtained when force is multiplied with the distance
As the force is conserved we have
The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.
Hope this helps! Please mark as brainliest!
Hello
1) First of all, since we know the radius of the wire (
), we can calculate its cross-sectional area
2) Then, we can calculate the current density J inside the wire. Since we know the current,
, and the area calculated at the previous step, we have
3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity
of the aluminium, the electric field is given by
Answer: The mass of the sculpture is 11.8kg
Explanation:
Using the equation of fundamental frequency of a taut string.
f = (1/2L)*√(T/μ) .... (Eqn1)
Where
f= frequency in Hertz =80Hz
T = Tension in the string = Mg
M represent the mass of the substance (sculpture) =?
g= 9.8m/s^2
L= Length of the string=90cm=0.9m
μ= mass density = mass of string /Length of string
mass of string =5g=0.005kg
L=0.9m
μ=0.005/0.9 = 0.0056kg/m
Using (Eqn1)
80= 1/(2*0.9) √(T/0.0056)
144= √(T/0.0056)
Square both sides
20736= T/0.0056
T= 116.12N
Recall that T =Mg
116.12= M * 9.8
M=116.12/9.8
M= 11.8kg
Therefore the mass of the sculpture is 11.8kg
To solve this problem, we use the formula
λ = s sin θ
where s is the separation and θ is the angle interference
So,
λ = 20 x 10^-6 sin 2.5
λ = 8.72 x 10^-7 m
The required angle for the fourth order bright fringe is
θb = sin⁻¹ (4λ / s) = sin⁻¹ (4 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 10.04°
The required angle for the fourth order dark fringe is
θd = sin⁻¹ (4.5 λ / s) = sin⁻¹ (4.5 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 11.31°
