Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S
= 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S
> σ
Answer:
The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
Explanation:
As data is incomplete here, so by seeing the complete question from the search the data is
vx_0=1.1 x 10^6
ax=0 As acceleration is zero in the horizontal axis so
Equation of motion in horizontal direction is given as


Now for the vertical distance
vy_o=0
than the equation of motion becomes

Now using this acceleration the value of electric field is calculated as

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
<span>(300 m/s)/(10 r/s) = 30 m/round.</span>
A solenoid hope this is right
Answer:
109.32 N/m
Explanation:
Given that
Mass of the hung object, m = 8 kg
Period of oscillation of object, T = 1.7 s
Force constant, k = ?
Recall that the period of oscillation of a Simple Harmonic Motion is given as
T = 2π √(m/k), where
T = period of oscillation
m = mass of object and
k = force constant if the spring
Since we are looking for the force constant, if we make "k" the subject of the formula, we have
k = 4π²m / T², now we go ahead to substitute our given values from the question
k = (4 * π² * 8) / 1.7²
k = 315.91 / 2.89
k = 109.32 N/m
Therefore, the force constant of the spring is 109.32 N/m