In 1967, John Wheeler invented the name "black hole" for the first time. However, there is no unified naming system for black holes.
Answer:
Every electric circuit in a wiring system must be protected against overloads. A circuit overload occurs when the amount of current flowing through the circuit exceeds the rating of the protective devices. The amount of current flowing in a circuit is determined by the load -- or the "demand" -- for current.
Explanation:
Hope this helps :)
Answer:
true
Explanation:
I did this unit for science
Answer:
3 times louder
Explanation:
The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².
Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹
and I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₁ = 10㏒(I₁/I₀)
L₁ = 10㏒(10⁻⁹/10⁻¹²)
L₁ = 10㏒(10³)
L₁ = 3 × 10㏒10
L₁ = 30㏒10
L₁ = 30 dB
Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₂ = 10㏒(I₁/I₀)
L₂ = 10㏒(10⁻³/10⁻¹²)
L₂ = 10㏒(10⁹)
L₂ = 9 × 10㏒10
L₂ =90㏒10
L₂ = 90 dB
So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3
So, Braylee's music is 3 times louder than Jessica's music
Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.
