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e-lub [12.9K]
2 years ago
8

Expansionary monetary policies would likely cause

Physics
1 answer:
Jlenok [28]2 years ago
6 0

Answer:

TRUE

Explanation:i've seen one and please make me the brainlest

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Sodium lamps give off light at
Ratling [72]

Answer:

θ = 17.67°

Explanation:

The grating equation can be used here to find the angle. The grating equation is given as follows:

m\lambda = dSin\ \theta

where,

m = order = 2

d = 3.88 x 10⁻⁶ m

λ = wavelength of light = 589 nm = 5.89 x 10⁻⁷ m

θ = angle = ?

Therefore, using these values in the equation, we get:

(2)(5.89\ x\ 10^{-7}\ m) = (3.88\ x\ 10^{-6}\ m)Sin\theta \\Sin\theta = \frac{(2)(5.89\ x\ 10^{-7}\ m)}{(3.88\ x\ 10^{-6}\ m)}\\\\\theta = Sin^{-1}(0.3036)

<u>θ = 17.67°</u>

5 0
3 years ago
Read 2 more answers
The speed of light in amber is
Umnica [9.8K]

Answer:

1.55

Explanation:

that it is the standard refractive index.

6 0
3 years ago
A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
Alex73 [517]

Answer:

F_{Earth}= 15.57 N

F_{Moon}= 2.60 N

F_{Uranus}= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

F_{Earth}=m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

F_{Uranus}=m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0
3 years ago
What is denser medium?​
abruzzese [7]

Answer: A medium in which speed of light is more is known as optically rarer medium and a medium in which speed of light is less is said to be optically denser medium. For example in air and water, air is raer and water is a denser medium.

Explanation:

6 0
3 years ago
How can an element be identified using its emission spectrum?
posledela
The correct choice is C .

Technically, choice-A is the process involved.  But it's already been done,
and widely published, and the spectrum of every element is easily available. 

For you to take the time and go to the trouble of constructing the spectra
would be like inventing wheels if you want to ride a bicycle. 
4 0
3 years ago
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