Answer:
BaF2
Explanation:
since you got the valence numbers just do the scissors move where you:
give the F the 2 and give the Ba the 1 so it be like
BaF2 here is the chemical compound
Answer:
92.33Hz
Explanation:
A tuba is considered as having one open end and the other end is closed. Then:
f_n = n×v / 4L
so here
Where n = 4
f_4 = 116.5 Hz
116.5 Hz = nv/4L
116.5 Hz = 4v/4L
4×v / 4L = v / L
116.5 Hz = v/L............1
Let's assume the speed of sound is 343 m/s. Then substituting 343 m/s for v in equation 1
116.5 Hz = v/L
116.5 Hz = 343 m/s/L
Making L the subject
116.5 Hz × L = 343 m/s
L = 343m/s / 116.5Hz
= 2.944 m
Add 0.721 m to the length, and
f_4 = 343m/s / (2.944+0.721)m
f_4 = 343m/s ÷ 3.715
= 92.328 Hz
Approximately = 92.33Hz
Hence the new frequency for the 4th harmonic is 92.33Hz
Answer:
so initial speed of the rock is 30.32 m/s
correct answer is b. 30.3 m/s
Explanation:
given data
h = 15.0m
v = 25m/s
weight of the rock m = 3.00N
solution
we use here work-energy theorem that is express as here
work = change in the kinetic energy ..............................1
so it can be written as
work = force × distance ...................2
and
KE is express as
K.E = 0.5 × m × v²
and it can be written as
F × d = 0.5 × m × (vf)² - (vi)² ......................3
here
m is mass and vi and vf is initial and final velocity
F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s
so put value in equation 3 we get
m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²
solve it we get
(vi)² = 919
vi = 30.32 m/s
so initial speed of the rock is 30.32 m/s