Answer:
perpendicular to the direction of the greatest stress
Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
Answer:
Secondary structure
The secondary structure arises from the hydrogen bonds formed between atoms of the polypeptide backbone. The hydrogen bonds form between the partially negative oxygen atom and the partially positive nitrogen atom
I can only offer you one which is mushroom. They're fungus' and they can't produce their own food. They get their food from decaying and dead animals and absorb it through their roots.
Answer:
v = 0
Explanation:
This problem can be solved by taking into account:
- The equation for the calculation of the period in a spring-masss system
( 1 )
- The equation for the velocity of a simple harmonic motion
( 2 )
where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block
Hence
and by reeplacing it in ( 2 ):
In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.
