Answer:
0.695s
Explanation:
From Hooke's law, the restoring force is given has
F = -ky .......1
Where F is the force, y is the spring displacement and k force constant of the spring.
Also recall,
F=mg ............ 2
Where m is the mass of object, g is the acceleration due to gravity.
Equating 1 and 2
Ky = mg
Given that g=9.8m/s2 , y is 3.4cm and g is 8g
K×3.4/100m =8/1000kg × 9.8m/s2
K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034
K= 0.0784÷0.035
K=2.24N/m
Mass ofvthe second object is 25g =0.025kg
Period of oscillation T
T=2π√m/k
T=2×3.142√0.025/2.24
T=6.284√0.0111
T=0.659seconds
Answer:
V = ω A sin ω t can be used to describe SHM
When sin ω t = 1 (the maximum value possible)
V = ω A at equilibrium sin ω t = 1
Your question is missing one part as "magnification"
i have completed the missing part below
Answer:
a. 
b. 
Explanation:
For this type of numerical we will use the following formulas
......... Eq1
where,
refractive index of the medium surrounding refracting surface/object
i.e. =
refractive index of the refracting surface/object
i.e. 
distance of object from the vertex of the refracting surface
distance of image from the vertex of the refracting surface
radius of curvature of the refracting surface
........... Eq2
where,
magnification
Convention:
Given:
=
because refraction surface is concave
Required:
a. 
b. 
Solution:
a. putting values in eq1, we get
cm
b. 
Answer:
volume of the airspace=0.605m^3
Explanation:
Patm = 10^5 N/m^2
Depth= 60 metres
Pressure at the depth = ?
Density = 1.025 g/cm3 = 1025 kg/m3
P = Patm + hσg
P = 10^5 + 60*1025*9.8
P = 702700 N/m^2
P = 7.027 atm
Since the temperature is constant, Boyle’s law holds
P1V1 = P2V2
P1 = 1 atm
P1, initial pressure of the bell(atm pressure) = 1 atm
The initial volume of the airspace, V1 = 4.25 m^3
Final pressure at the depth = 7.027 atm
Applying the Boyle’s lay
1*4.25 = 7.027 * V2
V2 = 4.25/7.027
V2 = 0.605m^3
