Answer: 340.8W
Explanation: Please see the attachments below
Respuesta: verifique amablemente la explicación
Explicación:
Dado lo siguiente:
Longitud (L) del cable = 120 m
Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m
Fuerza (F) = 380 N
Esfuerzo longitudinal = Fuerza / Área
Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4
Área = (3.142 * 4.84 * 10 ^ -6)
Área = 0.00000380132 m²
Estrés = Fuerza / Área
Estrés = 380 / 0.00000380132
Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2
Deformación longitudinal: extensión / longitud
Extensión = 0.10 m
Longitud = 120 m
Deformación longitudinal = 0,1 m / 120 m
Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
the speed after 3 seconds is 10 m/s
Explanation:
The computation of the speed is shown below:
As we know that
V = U + at
Here,
U = 34 m/s
a = - 8 m/s²
t = 3 Sec
V = velocity after 3 sec
V = 34 + (-8)3
= 34 - 24
V = 10 m/s
Hence, the speed after 3 seconds is 10 m/s
The lithosphere includes the brittle upper portion of the mantle and the crust, the outermost layers of Earth's structure. It is bounded by the atmosphere above and the asthenosphere (another part of the upper mantle) below. Although the rocks of the lithosphere are still considered elastic, they are not viscous
