Angular acceleration = (change in angular speed) / (time for the change)
Change in angular speed = (ending speed) minus (starting speed)
Change in angular speed = (16 rad/s) - (zero) = 16 rad/s .
Angular acceleration = (16 rad/s) / (0.4 s)
(Average) angular acceleration = 40 rad/s²
Answer:
The spring constant = 9.25 N/m
Explanation:
The equation of an object attached to a spring that is oscillating is
T = 2π√(m/k)
Where T = period of the oscillation, m = mass of the object, k = spring constant.
Making k the subject of the equation,
k = 4π²m/T²......................... Equation 1
Note: Period(T) is the time taken to complete one oscillation
Given: T = t/10 = 9.0/10 = 0.9 s, m = 190 g = 0.19 kg.
Constant: π = 3.14
Substitute these values into equation 1.
k = 4(3.14)²(0.19)/0.9²
k = 7.4933/0.81
k = 9.25 N/m
Thus the spring constant = 9.25 N/m
Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )
Answer:
R = 2Ω
Explanation:
Potential difference (V) = current (I) * Resistance (R)
V = IR
I = 2.0A
V = 10v
R = ?
V = IR
R = V / I
R = 10 / 2
R = 2Ω
The resistance across the wire is 2Ω
Answer:
Diameter will be 351.42 mm
Explanation:
We have given current flowing in the copper wire i = 310 A
Voltage drop across the wire V = 0.55 volt
We know that resistance is given by 
Length of the copper wire l = 1 m
Resisitivity of the copper wire 
We know that resistance 


As area 


So diameter
= 351.42 mm