A farsighted girl has a near point at 2.0 m but has forgotten her glasses at home. The girl borrows eyeglasses that have a power
of 2.75 diopters. These are not going to make her able to view normally. With these eyeglasses, what is the new near point of the girl, assuming that she wears them extremely close to her eyes
1 answer:
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Answer: The frequency = 1714.3Hz
Explanation: The solution can be achieved by using doppler effect formula.
Since the source is moving toward the observer, the velocity of the observer will be positive.
Please find the attached file for the solution
Answer:
v₀ₓ = 15 m / s, = 5.2 m / s
v = 15.87 m / s , θ = 19.1
Explanation:
This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed
Vy² =² - 2 g y
² =
² + 2 g y
= √ (
² + 2 gy
Let's calculate
= √ (1.25² + 2 9.8 1.3)
= √ (27.04)
= 5.2 m / s
The initial speed can be calculated by the initial speed
v = √ v₀ₓ² + ²
v = RA (15² + 5.2²)
v = 15.87 m / s
We look for the angle with trigonometry
tan θ = voy / vox
θ = tan⁻¹ I'm going / vox
θ = tan⁻¹ 5.2 / 15
θ = 19.1
The answer is
v₀ₓ = 15 m / s
= 5.2 m / s