Answer:
526.57 Pa
Explanation:
P ( pressure at the bottom of the container) = 1.049 × 10^5 pa
Using the formula of pressure in an open liquid
Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²
Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa
P( atmospheric pressure) = 1.013 × 10^5 Pa
Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g
Subtract each of the pressure from the absolute pressure at the bottom
P(bottom) - atmospheric pressure
(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa
subtract pressure due to water from the remainder
3600 - 2050.29 = 1549.71 Pa
1549.71 = ρ(density of the liquid) × h (depth of the liquid) × g
ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa
Answer:
(a) λ=77×10⁻⁶m
(b)Yes its enough to examine entire eye,
(c)λ=16.65μm.
Explanation:
The ultrasonic sound are used to detect problems in human body by using device consisting of transmitter and detector .
The speed of sound in human tissue is 1540m/s.
The wavelength of sound is 20MHz.
(a) let the velocity of sound in human tissue is v = 1540m/s.
v=fλ
λ= 
=
= 77×10⁻⁶m
λ=77×10⁻⁶m
(b) we can scan effectively upto the depth of about 500λ
500λ=500× (77×10⁻⁶)= 3.85cm
Yes its enough to examine entire eye,
(c) In air the velocity of sound is 331m/s.
λ= = 
= 16.65μm
λ=16.65μm.
Correct answer choice is :
A) Energy moves from one place to another through the wave.
Explanation:
Waves can carry energy over a range without moving matter the complete distance. For instance, an ocean wave can move many kilometers without the water itself traveling many kilometers. The water moves up and down a motion recognized as a disturbance. A wave brings its energy without carrying matter. Waves are observed to travel through an ocean or lake, yet the water perpetually reverts to its rest state. Energy is carried through the medium, yet the water particles are not carried.
Answer:
Magnitude of the net force acting on the kayak = 39.61 N
Explanation:
Considering motion of kayak:-
Initial velocity, u = 0 m/s
Distance , s = 0.40 m
Final velocity, v = 0.65 m/s
We have equation of motion v² = u² + 2as
Substituting
v² = u² + 2as
0.65² = 0² + 2 x a x 0.4
a = 0.53 m/s²
We have force, F = ma
Mass, m = 75 kg
F = ma = 75 x 0.53 = 39.61 N
Magnitude of the net force acting on the kayak = 39.61 N
