Easy !
Take any musical instrument with strings ... a violin, a guitar, etc.
The length of the vibrating part of the strings doesn't change ...
it's the distance from the 'bridge' to the 'nut'.
Pluck any string. Then, slightly twist the tuning peg for that string,
and pluck the string again.
Twisting the peg only changed the string's tension; the length
couldn't change.
-- If you twisted the peg in the direction that made the string slightly
tighter, then your second pluck had a higher pitch than your first one.
-- If you twisted the peg in the direction that made the string slightly
looser, then your second pluck had a lower pitch than the first one.
Echoes occur when a reflected soundwave reaches the ear more than 0.1 seconds after the original sound wave was heard. ... There will be an echoinstead of a reverberation. Reflection of sound waves off of surfaces is also affected by the shape of the surface.
(mark as brainly please)
Answer:
P max = 1000 pa
P min = 200 pa
Explanation:
P = F/A
pressure will be maximum when surface gets minimum. so to find the maximum amount of pressure we need to calculate the minimum surface. it is 2cm×5cm = 10cm² = 0.001m² . then we have:
P = 1 / 0.001 = 1000 pa
to find minimum pressure the surface that must be chosen is 10cm×5cm = 50cm² = 0.005m² .
P = 1 / 0.005 = 200 pa
Answer:
p= 1.50289×10⁷ N/m²
Explanation:
Given
HA = (564 m)................(River Elevation)
HB = (2096 m).............(Village Elevation)
Area = A =(π/4){Diameter}² = (π/4){0.15 m}² = 0.017671 m²
ρ = (1 gram/cm³) = (1000 kg/m³)........(Water Density)
p(pressure)=?
Solution
p=PA - PB
p= ρ*g*HB - ρ*g*HA
p= (ρ*g)*(HB - HA)
p= (1000×9.81 )×{2096 - 564}
p= 1.50289×10⁷ N/m²
