Answer:
Thorium 227 (also known as Radioactinium)
Answer:
2 seconds
Explanation:
The frequency of a wave is related to its wavelength and speed by the equation

where
f is the frequency
v is the speed of the wave
is the wavelength
For the wave in this problem,
v = 2 m/s

So the frequency is

The period of a wave is equal to the reciprocal of the frequency, so for this wave:

This means that the wave takes 4 seconds to complete one full cycle.
Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

Answer:
Explanation:
Let the volume below water be v . Then
buoyant force = v d g where d is density of water , g is acceleration due to gravity
= v x 1000 x g
weight of wood piece = volume x density of wood x g
= .6 x 600 x g
for equilibrium while floating
buoyant force = weight
= v x 1000 x g = .6 x 600 x g
v = .36 m²
volume above water or volume exposed = .6 - .36
= .24 m²
When immersed completely ,
buoyant force = .6 x 1000 x 9.8
= 5880 N
weight of wood
= .6 x 600 x g
= 3528 N
buoyant force is more than the weight . In order to equalise them for floating with full volume in water
weight required = 5880 - 3528
= 2352 N.
Explanation:
Given that,
Diameter = 10 cm
Distance = 2 m
Speed 
Speed 
Pressure in main pipe
(I). We need to calculate the diameter
Using equation of continuity





(II). We need to calculate the pressure the gauge pressure
Using Bernoulli equation




(III). If it is possible to carry water to a faucet 17 m above ground,
Using Bernoulli equation


Here, 
Put the value in the equation


Hence, This is required solution.