Answer:
1.16cm were cut off the end of the second pipe
Explanation:
The fundamental frequency in the first pipe is,
<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>
f1 = v/4L, where v is the speed of sound and L is the length of the pipe
266 = 340/4L
L = 0.31954 m = 0.32 m
It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.
<u>So, the length of the second pipe is L – L’</u>
Then, the fundamental frequency in the second pipe is
f2 = v/4(L - L’)
<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>
f2 – f1 = 10hz
[v/4(L - L’)] – 266 = 10
[v/4(L – L’)] = 10 + 266
[v/4(L – L’)] = 276
(L - L’) = v/(4 x 276)
(L – L’) = 340/(4 x 276)
(L – L’) = 0.30797
L’ = 0.31954 – 0.30797
L’ = 0.01157 m = 1.157 cm ≅ 1.16cm
Hence, 1.16 cm were cut from the end of the second pipe
Here, F = m * a
F = m * v/t
Here, m = 81 Kg
v = 22 m/s
t = 1,4 s
Substitute their values,
F = 81 * 22/1.4
F = 81 * 15.71
F = 1273 N
So, Closest value from your options is 1300 N
In short, Your Answer would be Option B
Hope this helps!
Answer:
Coefficient of kinetic friction = 0.146
Explanation:
Given:
Mass of sled (m) = 18 kg
Horizontal force (F) = 30 N
FInal speed (v) = 2 m/s
Distance (s) = 8.5 m
Find:
Coefficient of kinetic friction.
Computation:
Initial speed (u) = 0 m/s
v² - u² = 2as
2(8.5)a = 2² - 0²
a = 0.2352 m/s²
Nweton's law of :
F (net) = ma
30N - μf = 18 (0.2352)
30 - 4.2336 = μ(mg)
25.7664 = μ(18)(9.8)
μ = 0.146
Coefficient of kinetic friction = 0.146
Answer:
I can't get a clear explanation of the question
The correct answer for this question is this one: "C. Neither Natalie nor Will." Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the people in the household. <span>Will says that socialization is easily attained if the animal is first exposed to humans after 12 weeks of age.</span>
