Answer:
a) factor 
b) factor 
c) factor 
d) factor 
Explanation:
Time period of oscillating spring-mass system is given as:


where:
frequency of oscillation
mass of the object attached to the spring
stiffness constant of the spring
a) <u>On doubling the mass:</u>
- New mass,

<u>Then the new time period:</u>




where the factor
as asked in the question.
b) On quadrupling the stiffness constant while other factors are constant:
New stiffness constant, 
<u>Then the new time period:</u>

where the factor
as asked in the question.
c) On quadrupling the stiffness constant as well as mass:
New stiffness constant, 
New mas, 
<u>Then the new time period:</u>

where factor
as asked in the question.
d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.
so, factor 
The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer:
Explanation:
40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.
Answer:
The current is 2.0 A.
(A) is correct option.
Explanation:
Given that,
Length = 150 m
Radius = 0.15 mm
Current density
We need to calculate the current
Using formula of current density


Where, J = current density
A = area
I = current
Put the value into the formula


Hence, The current is 2.0 A.
Answer:
Total energy saving will be 0.8 KWH
Explanation:
We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW
30 bulbs are of power 60 W
So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW
Total power of 80 bulbs = 1.8+5 = 6.8 KW
Total time = 3 hour
We know that energy 
Now power of each CFL bulb = 25 W
So power of 80 bulbs = 80×25 = 2000 W = 2 KW
Energy of 80 bulbs = 2×3 = 6 KWH
So total energy saving = 6.8-6 = 0.8 KWH