It is an example of reflection of wave. C.
Answer:
Δ L = 2.57 x 10⁻⁵ m
Explanation:
given,
cross sectional area = 1.6 m²
Mass of column = 26600 Kg
Elastic modulus, E = 5 x 10¹⁰ N/m²
height = 7.9 m
Weight of the column = 26600 x 9.8
= 260680 N
we know,
Young's modulus=
stress = 
= 
= 162925
strain = 
now,



Δ L = 2.57 x 10⁻⁵ m
The column is shortened by Δ L = 2.57 x 10⁻⁵ m
Answer:
4.25 J
Explanation:
Given that
mass of plastic ball = 11 g
Mass of plastic ball = 0.011 kg
velocity of ball = 29 m/s
We know that from work power energy theorem

We know that kinetic energy of moving mass given as

Now by pitting the values


KE= 4.25 J
So the work done on the ball is 4.25 J
Answer:
C 80 m
Explanation:
Given:
v₀ = 30 m/s
a = -10 m/s²
t = 8 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²
Δy = -80 m
The ball lands 80 m below where it started. So the height of the cliff is 80 m.
The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J
Assuming air friction is negligible,
a = - 9.8 m / s²
u = 31.4 m / s
s = 30 m
v² = u² + 2 a s
v² = 31.4² + ( 2 * - 9.8 * 30 )
v² = 985.96 - 588
v² = 397.96 m / s
KE = 1 / 2 m v²
KE = 1 / 2 * 0.155 * 397.96
KE = 0.0775 * 397.96
KE = 30.85 J
Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J
To know more about kinetic energy
brainly.com/question/24360064
#SPJ1