The appropriate response is letter D. The wave ventures slower and with an expanded wavelength when a sound wave entering a range of hotter air. Hotter air implies less thick, so the wave ought to back off.
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Answer:
block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s
Explanation:
We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.
Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)
Before the crash
p₀ = m v₀ + 0
After the crash
= (m + M) v
p₀ =
m v₀ = (m + M) v (1)
Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring
Initial
Em₀ = K = ½ m v2
Final
E = Ke = ½ k x2
Emo = E
½ m v² = ½ k x²
v² = k/m x²
Let's look for the spring constant (k), with Hook's law
F = -k x
k = -F / x
k = - 0.75 / -0.25
k = 3 N / m
Let's calculate the speed
v = √(k/m) x
v = √ (3/8.00) 0.15
v = 0.09186 = 9.18 10⁻² m/s
This is the spped of the block plus bullet rsystem right after the crash
We substitute calculate in equation (1)
m v₀ = (m + M) v
v₀ = v (m + M) / m
v₀ = 0.09186 (0.008 + 0.992) /0.008
v₀ = 11.5 m / s
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Answer:
1.5X10^-4C
Explanation:
Expression for the electric force between the two charges is given by -
F = (k*q1*q2) / r^2
Here, k = constant = 9 x 10^9 N*m^2 / C^2
q1 = unknown
F=25N
q2 = 1.9x10^-6 C
r = 0.32 m
Substitute the given values in the above expression -
25=9x10^9 *(q1) * 1.9 x10^ -6/ (0.32m)^2
25= 17,100* (q1)/ 0.1024
multiply both sides by 0.1024 to get rid of the denominator
2.56=17,100*(q1)
Divide both sides by 17,100 to isolate q1
q1=1.5x10^-4C
Answer:
fog and mist are the examples