Answer:
i)ω=3600 rad/s
ii)V=7059.44 m/s
iii)F=1245.8 N
Explanation:
i)
We know that angular speed given as

We know that for one revolution
θ=2π
Given that time t= 2 hr
So
ω=θ/t
ω=2π/2 = π rad/hr
ω=3600 rad/s
ii)
Average speed V

Where M is the mass of earth.
R is the distance
G is the constant.
Now by putting the values


V=7059.44 m/s
iii)
We know that centripetal fore given as

Here given that m= 200 kg
R= 8000 km
so now by putting the values


F=1245.8 N
Answer:
(a) T = W/2(1-tanθ) (b) 39.81°
Explanation:
(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:
Summation of moment in clockwise direction is equivalent to zero. Therefore,
T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0
T*l*(cosθ - sinθ) = W*(l/2)*cosθ
T = W*cosθ/2(cosθ - sinθ)
Dividing both the numerator and denominator by cosθ, we have:
T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)
(b) If T = 3W, then:
3W = W/2(1-tanθ),
Further simplification and rearrangement lead to:
1 - tanθ = 1/6
tanθ = 1 - (1/6) = 5/6
θ = tan^(-1) 5/6 = 39.81°
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Answer:
modulus =3.97X10^6 Ib/in^2, Poisson's ratio = 0.048
Explanation:
Modulus is the ratio of tensile stress to tensile strain
Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain within the direction of the stretching force
And contraction occur from 0.6 in x 0.6 in to 0.599 in x 0.599 in while 2 in extended to 2.007, with extension of 0.007 in
Answer:
Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)
where
σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa
α = crack length = 3.8 × 10–2 mm = 3.8 x 10⁻⁵m
ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m
Substituting these values into the formula, we can calculate the max stress as
====== 2 x 1.40x 10⁸ x √(3.8 x 10⁻⁵/1.9 × 10⁻⁷)
σₐ = 24.4MPa