Answer:
The magnification is -6.05.
Explanation:
Given that,
Focal length = 34 cm
Distance of the image =2.4 m = 240 cm
We need to calculate the distance of the object
Where, u = distance of the object
v = distance of the image
f = focal length
Put the value into the formula
The magnification is
Hence, The magnification is -6.05.
Answer:
0.191 s
Explanation:
The distance from the center of the cube to the upper corner is r = d/√2.
When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.
Sum of the torques:
∑τ = Iα
Fr cos θ = Iα
(k r sin θ) r cos θ = Iα
kr² sin θ cos θ = Iα
k (d²/2) sin θ cos θ = Iα
For a cube rotating about its center, I = ⅙ md².
k (d²/2) sin θ cos θ = ⅙ md² α
3k sin θ cos θ = mα
3/2 k sin(2θ) = mα
For small values of θ, sin θ ≈ θ.
3/2 k (2θ) = mα
α = (3k/m) θ
d²θ/dt² = (3k/m) θ
For this differential equation, the coefficient is the square of the angular frequency, ω².
ω² = 3k/m
ω = √(3k/m)
The period is:
T = 2π / ω
T = 2π √(m/(3k))
Given m = 2.50 kg and k = 900 N/m:
T = 2π √(2.50 kg / (3 × 900 N/m))
T = 0.191 s
The period is 0.191 seconds.
Answer:
A. it's the only answer that makes sense. if I'm wrong sorry
Answer:
0.74 N/cm
Explanation:
The following data were obtained from the question:
Mass (m) = 3 Kg
Extention (e) = 40 cm
Spring constant (K) =?
Next, we shall determine the force exerted on the spring.
This can be obtained as follow:
Mass (m) = 3 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 3 × 9.8
F = 29.4 N
Finally, we shall determine the spring constant of the spring. This can be obtained as follow:
Extention (e) = 40 cm
Force (F) = 29.4 N
Spring constant (K) =?
F = Ke
29.4 = K × 40
Divide both side by 40
K = 29.4 / 40
K = 0.74 N/cm
Therefore, the spring constant of the spring is 0.74 N/cm
A higher frequency. I just got this answer on usa test prep.
