I looked it up and it gave me educational exercise, but I don't know if it is right.
Answer:
T=1022.42 N
Explanation:
Given that
l = 32 cm ,μ = 1.5 g/cm
L =2 m ,V= 344 m/s
The pipe is closed so n= 3 ,for first over tone
f= 129 Hz
The tension in the string given as
T = f²(4l²) μ
Now by putting the values
T = f²(4l²) μ
T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100
T=1022.42 N
1.41 × 10³⁰ MeV
As we know, E=mc², where E is energy, m is mass and c is the speed of light(i.e. 3×10⁸ m/s).
Given mass = 2.5 kgs
∴ E = (2.5)×(3×10⁸)² J = 22.5×10¹⁶ J
As our answer is in joules so we have to convert it into mega electron volt(MeV)
1 J = 6.242 × 10¹² MeV
⇒ 22.5×10¹⁶ J = 22.5×10¹⁶ × (6.242 × 10¹²) MeV
⇒1.41 × 10³⁰ MeV
If you want to learn more about mass-energy conversions then you can check out this link:
https://brainly.in/question/9760064
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Temperature: Kelvin or degree Celsius; thermometer
Length: meter and its fractions and multiples; ruler
Volume: liter or cubic meter, mm, cm, km etc.; ruler for regular solids or empty spaces, graduated cylinder or kitchen measuring cup for liquids and irregular solids
Mass: kilogram and its multiples and fractions; balance with calibrated samples, or scale with knowledge of local gravity, or methods of applying known force and measuring acceleration
