I believe it is C. remain the same. If i am wrong i sincerely apologize.
Answer:
6 cm long
Explanation:
F = 4110N
Vo(speed of sound) = 344m/s
Mass = 7.25g = 0.00725kg
L = 62.0cm = 0.62m
Speed of a wave in string is
V = √(F / μ)
V = speed of the wave
F = force of tension acting on the string
μ = mass per unit density
F(n) = n (v / 2L)
L = string length
μ = mass / length
μ = 0.00725 / 0.62
μ = 0.0116 ≅ 0.0117kg/m
V = √(F / μ)
V = √(4110 / 0.0117)
v = 592.69m/s
Second overtone n = 3 since it's the third harmonic
F(n) = n * (v / 2L)
F₃ = 3 * [592.69 / (2 * 0.62)
F₃ = 1778.07 / 1.24 = 1433.927Hz
The frequency for standing wave in a stopped pipe
f = n (v / 4L)
Since it's the first fundamental, n = 1
1433.93 = 344 / 4L
4L = 344 / 1433.93
4L = 0.2399
L = 0.0599
L = 0.06cm
L = 6cm
The pipe should be 6 cm long
Answer:
11.) g = 3.695 m/s^2
12.) g = 8.879 m/s^2
13.) E = 8127 N/C
Explanation:
11.) Given that the
Mercury mass M = 3.3 × 10^23kg
Radius r = 2.44 ×10^6 m
Gravitational constant G = 6.67408 × 10^-11 m3kg-1 s^-2
Gravitational field strength g can be calculated by using the formula below
g = GM/r^2
Substitutes all the parameters into the formula
g = (6.67408 × 10^-11 × 3.3 × 10^23)/(2.44×10^6)^2
g = 2.2×10^13/5.954×10^12
g = 3.695 m/s^2
12.) Given that the
Venus mass M = 4.87×10^24kg
Radius r = 6.05 × 10^6 m
Using the same formula for gravitational field strength g
g = GM/R2
Substitute all the parameters into the formula
g = (6.67408 × 10^-11 × 4.87×10^24)/(6.05×10^6)^2
g = 3.25×10^14/3.66×10^13
g = 8.879 m/s^2
13.) Given that the
Charge = 2.26 nC = 2.26×10^-9
Distance d = 0.05m
Electric field strength E can be calculated by using the formula below
E = Kq/d^2
Where
K = electrostatic constant 8.99 × 10^9 Nm2/C2
Substitutes all the parameters into the formula
E = (8.99 × 10^9 × 2.26×10^-9)/0.05^2
E = 20.3174/2.5×10^-3
E = 8126.96 N/C
The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.
F α Qq
If the charges are now half the initial charges:
<span>F α (1/2)Q *(1/2)q
</span>
F α (1/4)Q<span>q
The new force when the charges are each halved is (1/4) the first initial force experienced at full charge.</span>
