Answer:
x' = 1.01 m
Explanation:
given,
mass suspended on the spring, m = 0.40 Kg
stretches to distance, x = 10 cm = 0. 1 m
now,
we know
m g = k x
where k is spring constant
0.4 x 9.8 = k x 0.1
k = 39.2 N/m
now, when second mass is attached to the spring work is equal to 20 J
work done by the spring is equal to
x'² = 1.0204
x' = 1.01 m
hence, the spring is stretched to 1.01 m from the second mass.
Answer:
1450.4 KN
Explanation:
Pressure = ρhg
where: ρ is the density of the liquid, h is the height and g the force of gravity.
Total pressure exerted by the liquids at the base = Pressure of oil + Pressure of water + Pressure of mercury
So that,
i. Pressure of oil = ρhg
(ρ = 0.8 g/cm³ = 800 kg/m³)
= 800 x 5 x 9.8
= 39200
Pressure of oil = 39200 N
ii. Pressure of water = ρhg
(ρ = 1 g/cm³ = 1000 kg/m³)
= 1000 x 8 x 9.8
= 78400
Pressure of water = 78400 N
ii. Pressure of mercury = ρhg
(ρ = 13.6 g/cm³ = 13600 kg/m³)
= 13600 x 10 x 9.8
= 1332800
Pressure of mercury = 1332800 N
So that,
Total pressure exerted by the liquids at the base = 39200 + 78400 + 1332800
= 1450400
= 1450.4 KN
Total pressure exerted by the liquids at the base is 1450.4 KN.
Answer:
1.07 m
Explanation:
x = Compression of the spring
k = Spring constant = 53 N/m
Initial length = 18 cm
P = Kinetic energy
K = Kinetic energy
At the lowest point of the mass the energy conservation is as follows
At its lowest position the mark on the ruler will be
The spring line will end up at 1.07 m