To solve this problem we will apply the concepts related to resonance. The velocity with which sound travels in any medium may be determined if the frequency and the wavelength are known. Since the pipe is closed at one end, that produces frequencies ratio 1:3:5, then
Third resonance occurs at 
At resonance 
Therefore at 1m will the tuning fork again create a resonance condition
Answer:
b= 2.14 m
Explanation:
Given that
Weight of the board ,wt = 40 N
Wight of the first children , wt₁=500 N
Weight of the second children ,wt₂ = 350 N
The distance of the 500 N child from center ,a= 1.5 m
lets take distance of the 350 N child from center = b m
Now by taking the moment about the center of the board
We know that moment = Force x Perpendicular distance from the force
wt₁ x a = wt₂ x b
500 x 1.5 = 350 x b
b= 2.14 m
Therefore the distance of the 350 N weight child from the center is 2.14 m.
Answer:
a) P = 15993.6 N/m²
b) P' = 11062.24 N/m²
Explanation:
Given:
Density of the mercury, ρ = 13.6 × 10³ kg/m³
The blood pressure reading is given as 120 over 83 mm
a) Now,
The pressure (P) due to a fluid is given as:
P = ρgh
where, g is the acceleration due to the gravity
h is the pressure head
The systolic pressure head from the blood pressure reading is , h = 120 mm = 0.12 m
substituting the value in the formula for pressure,
we get
P = 13.6 × 10³ × 9.8 × 0.12
or
P = 15993.6 N/m²
b) The systolic pressure head from the blood pressure reading is , h' = 83 mm = 0.083 m
substituting the value in the formula for pressure,
we get
P' = 13.6 × 10³ × 9.8 × 0.083
or
P' = 11062.24 N/m²
Answer:
Explanation:
Given:
- gravitational field strength of moon at a distance R from its center,
- Distance of the satellite from the center of the moon,
<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>
∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.
