The velocity of the target and arrow after collision is 6.67m/s
<u>Explanation:</u>
Given:
Mass of arrow, mₐ = 415g
Speed of arrow, vₐ = 68.5m/s
Mass of the target, mₓ = 3.3kg = 3300g
speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction
Velocity of the target and arrow after collision, vₙ = ?
Applying the conservation of momentum,
mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ
415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ
28427.5 - 3630 = 3715 X vₙ
24797.5 = 3715 X vₙ
vₙ = 6.67m/s
Therefore, the velocity of the target and arrow after collision is 6.67m/s
Answer: Please see answer in explanation column.
Explanation:
Given that
v≈(331 + 0.60T)m/s
where Temperature, T = 14°C
v≈(331 + 0.60 x 14)m/s
v =331+ 8.4 = 339.4m/s
In our solvings, note that
f= frequency
λ=wavelength
L = length
v= speed of sound
a) Length of the pipe is calculated using the fundamental frequency formulae that
f=v/2L
Length = v/ 2f
= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m
b) wavelength of the fundamental standing wave in the pipe
L = nλ/2,
λ = 2L/ n
λ( wavelength )= 2 x 0.3435/ 1
= 0.687m
c) frequency of the fundamental standing wave in the pipe
F = v/ λ
= 339.4m/s/0.687m=
494.03s^-1 = 494 Hz
d) the frequency in the traveling sound wave produced in the outside air.
This is the same as the frequency in the open organ pipe = 494Hz
e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m
f) To play D above middle c . the distance is given by
L =v/ 2 f
= 343/ 2 x 294
=0.583m
Answer:
C
Explanation:
The strong force holds together quarks, the fundamental particles that make up the protons and neutrons of the atomic nucleus, and further holds together protons and neutrons to form atomic nuclei. As such it is responsible for the underlying stability of matter.
E all of the answers above correlate to the student and his skateboard
Answer: chemical property
Explanation:
