Answer:
B. d(low)=4d(high)
Explanation:
Frequency of a string can be written as;
f = v/2L
Where;
v = sound velocity
L = string length
Frequency can be further expanded to;
f = v/2L = (1/2L)√(T/u) ......1
Where;
m= mass,
u = linear density of string,
T = tension
p = density of string material
A = cross sectional area of string
d = string diameter
u = m/L .......2
m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)
f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))
f = (1/2L)√(T/((p(πd^2)L/4)/L))
f = (1/2L)√(4T/pπd^2)
f = (1/L)(1/d)√(4T/pπ)
Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.
f ~ 1/d
So, if
4f(low) = f(high)
Then,
d(low) = 4d(high)
Answer:
tbh vector does not have any direction at all the answer is 0
Answer:
whats the Question? (i play softball and its amazing btw lol)(ugh staying at home is getting to me)
Explanation:
Answer:
1.603 s
Explanation:
Given that
Initial mass, = 0.45 kg
Initial period, = 1.45 s
Initial radius, = 0.14 m
Final mass, = 0.55 kg
Final period, = ?
Final radios, = 0.14 m
Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that
m₁r₁ω₁² = m₂r₂ω2²
Where, ω = 2π/T, on substituting, we have
0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²
0.45 / 1.45² = 0.550 / T₂²
T₂² = 0.550 * 1.45² / 0.45
T₂² = 2.56972
T₂ = √2.56972
T₂ = 1.603 sec
