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Taya2010 [7]
3 years ago
12

You are using a diffraction grating with 2230 lines per centimeter. White light passes through it, and the grating spreads the l

ight out into its spectral components. At what angle does green light of wavelength 508 nm appear in second order
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

The angle of diffraction is 13.09°

Explanation:

Given:

Wavelength of light \lambda = 508 \times 10^{-9} m

Order of diffraction n = 2

No. lines per centimeter N = 2230

First find the grating element,

  d = \frac{1}{N}

  d = 4.484 \times 10^{-6} m

From the formula of diffraction,

  d \sin \theta = n \lambda

    \sin \theta = \frac{2 \lambda}{d}

    \sin \theta = \frac{2 \times 508 \times 10^{-9} }{4.484 \times 10^{-6} }

    \theta = \sin ^{-1} (0.2265)

    \theta = 13.09°

Therefore, the angle of diffraction is 13.09°

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An object slides along a curved track of negligible friction, as shown in the figure. The potential energy U of the object as a
balandron [24]

Answer:

E) The object achieves its maximum speed at approximately x=3m.

Explanation:

Since the object is moving in gravitational field

potential energy + kinetic energy = constant = 14 J

At  x = 3 m , potential energy is minimum that is zero

kinetic energy will be maximum .

Kinetic energy = 14 - 0 = 14 J

Hence option ( E ) is the answer.

5 0
3 years ago
What four things<br> are required for photosynthesis?
drek231 [11]

Explanation:

carbon dioxide, chlorophyll, water and light

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3 years ago
A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that
gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

v(t)=At+Bt^2

Differentiating with respect to time

\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt

At t = 0

1.4=A

Hence, A = 1.4 m/s²

B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3

B = -0.10493 m/s³

At t = 5 seconds

a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2

a = 1.29507 m/s²

T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N

T = 28095.8271 N

Weight of rocket

W=2530\times 9.81=24819.9\ N

\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W

T = 1.13198 W

3 0
4 years ago
A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 k
Bad White [126]

Answer:

v_{f,w} = 1.791\,\frac{m}{s}, v_{f,c} = 0.972\,\frac{m}{s}

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}

The final angular speed is:

\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}

\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev}  \right)

\omega_{f} \approx 5.116\,\frac{rad}{s}

The tangential velocities of the wheel and the clay are, respectively:

v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )

v_{f,w} = 1.791\,\frac{m}{s}

v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )

v_{f,c} = 0.972\,\frac{m}{s}

5 0
3 years ago
What is the relationship between distance of the objects and the gravitational force between them? (Another way of asking: What
Nadya [2.5K]

Explanation:

increase in the force of gravity

6 0
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