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kaheart [24]
3 years ago
7

Chadwick received the Doctor of? 1.Human Letters

Physics
1 answer:
cupoosta [38]3 years ago
7 0
I think it’s maybe 2. But I’m not sure
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What is night vision pls explain in long answers use<br> can use google
bulgar [2K]

Answer:

Night vision is the ability to see in low-light conditions. Whether by biological or technological means, night vision is made possible by a combination of two approaches: sufficient spectral range, and sufficient intensity range.

8 0
3 years ago
Radio waves travel at 3.00 · 108 m/s. Calculate the wavelength of a radio wave of frequency 900 kHz. (9.00 · 105 Hz.)
Anuta_ua [19.1K]
V: velocity of wave
f: frequency 
L: wavelenght

v = fL => L = v/f => L = (3x10^8)/(900x10^3) => L = 3.33 x 10^2m
7 0
3 years ago
An hydrogen molecule consists of two hydrogen atoms whose total mass is 3.3×10−27 kg and whose moment of inertia about an axis p
dlinn [17]

Answer:

6.9631\times 10^{-11}\ m

Explanation:

I = Moment of inertia = 4\times 10^{-48}\ kg m^2

m = Mass of two atoms = 2m = 3.3\times 10^{-27}\ kg

r  = distance between axis and rotation mass

Moment of inertia of the system is given by

I=mr^2\\\Rightarrow I=2mr^2\\\Rightarrow 4\times 10^{-48}=3.3\times 10^{-27}\times r^2\\\Rightarrow r=\sqrt{\frac{4\times 10^{-48}}{3.3\times 10^{-27}}}\\\Rightarrow r=3.48155\times 10^{-11}\ m

The distance between the atoms will be two times the distance between axis and rotation mass.

d=2r\\\Rightarrow d=2\times 3.48155\times 10^{-11}\\\Rightarrow d=6.9631\times 10^{-11}\ m

Therefore, the distance between the two atoms is 6.9631\times 10^{-11}\ m

3 0
3 years ago
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener
Elina [12.6K]

Answer:

Explanation:

From the given information:

The initial PE (PE)_i = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = P.E_f -P.E_i

ΔP.E = 0 - (PE)_i

ΔP.E = -P.E_i

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

\Delta P.E + \Delta K.E + \Delta U = 0

\Delta U = -\Delta P.E - \Delta K.E

this can be re-written as:

\Delta U =- (-\Delta P.E_i) - \Delta K.E

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

\Delta U =(70\%) \Delta P.E_i-0

\Delta U =(0.70) (490.5)

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

7 0
3 years ago
The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t
hram777 [196]

Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

(a) The relationship between the focal length and the focal length is given by:

f=\dfrac{1}{P}

f=\dfrac{1}{45.4}

f = 0.022 m

or

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Here, u = infinity

\dfrac{1}{v}=\dfrac{1}{2.2}

v = 2.2 cm

So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

5 0
4 years ago
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