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Musya8 [376]
3 years ago
11

Help me with number 3 and 4 pleaseeeeeeeee

Physics
2 answers:
Andrej [43]3 years ago
6 0

i think its b cause if you were talking about ice it freezes the melts the evaporates to gas

Roman55 [17]3 years ago
3 0
Number three I think might be b
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A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen sp
Colt1911 [192]

Answer:

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

           d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

           tan θ = y / L

           tan θ = \frac{sin \theta}{cos \theta} = sin θ

           sin θ = y / L

we substitute

          d \  \frac{y}{L} = m λ

with the initial data we look for the distance between the lines

           d = \frac{m \lambda \ L}{y}

           d = 1 656 10⁻⁹ 1.00 / 0.600

            d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

           λ = d y / L m

           λ = 1.09 10⁻⁶ 0.364 / 1.00 1

           λ = 3.9676 10⁻⁷ m

           λ = 3.967 10⁻⁷ m

         

we reduce nm

           λ = 396.7 nm

8 0
2 years ago
How do exons and introns differ?
Norma-Jean [14]
Exons And Introns differ because Exons code for protein and Introns do not.

so your answer would be: Exons code for Proteins



Are you in K12?
7 0
2 years ago
Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to
Leviafan [203]

Answer:

The centripetal acceleration of the object is 31550.72\ m/s^2.  

Explanation:

We have,

Radius of a circular path is 200 m

It takes 5 seconds to complete 10 revolutions. The angular velocity of the object is given by the rate of change of angular displacement per unit time :

\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s

The centripetal acceleration of the object is given by :

a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2

So, the centripetal acceleration of the object is 31550.72\ m/s^2.

6 0
3 years ago
A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?
Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

H_m = \frac{u^2sin^2 \theta}{2g}

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta  = sin^{-1}(0.2123)\\\\\theta  = 12.26^0

Therefore, the angle of projection is 12.26⁰.

6 0
2 years ago
This diagram shows the forces acting on a car. The center dot represents the car, a the arrows represent the forces acting on th
mylen [45]
Need diagram right? Post it
5 0
3 years ago
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