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3 years ago

Why are plants generally green?

Physics

1 answer:

e-lub [12.9K]3 years ago

7 0

The correct answer is option D, "The pigment chlorophyll reflects green light waves"

The chloroplast with in the plants consists of chlorophyll (green colour pigment) which is highly absorbent to the long and short wavelengths such as red and blue wavelength of visible spectrum and reflects green wavelengths thereby producing green colour visualizations to the viewer. The two common chlorophyll usually found in the leaves are chlorophyll A and B. Chlorophyll A is majorly responsible for reflecting the green light

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HELPPP MEEE PLEASSSE ASAP IT WOULD MEAN ALOT

777dan777 [17]

C. Lenses form images byvrefracting light; mirrors do not

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3 years ago

You push on a box with a force of 300 N directly north. Another person pushes the box with a

tangare [24]

Answer:

$resultant \\ \: F = \sqrt{ {300}^{2} + {600}^{2} } \\ = \sqrt{450000} \\ = 670.82 \: newtons$

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3 years ago

A 98.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.69 r

sertanlavr [38]

Answer:

Explanation:

The problem is based on conservation of angular momentum.

Moment of inertia of the disc = 1/2 m R² , m is mass of the disc and R is its radius.

= 1/2 x 98.1 x 1.51²

= 111.84 kg m²

Moment of inertia of disc + moment of inertia of bananas + monkey

= 1/2 x 98.1 x 1.51² + 9.29 x .45 x 1.51 + 20.3 x 1.51² ( moment of inertia of banana and monkey will be equal to mass x radial distance from axis² )

= 111.84 + 6.31 +46.28

= 164.43 kg m²

Now applying law of conservation of angular momentum

= I₁ ω₁ = I₂ω₂

111.84 x 1.69 = 164.43 x ω₂

ω₂ = 1.15 rad / s

6 0

3 years ago

Am I right? be honest

Bas_tet [7]

Answer:

I chose c because it is the greater slope at point c

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3 years ago

A 1000 Kg car approaches an intersection traveling north at 30 m/s . A 1250 Kg car approaches the same intersection traveling ea

Feliz [49]

Answer:

The final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

Explanation:

Taking north direction as positive y axis and east direction as positive x axis .

Given:

Mass of first car is, $m_1=1000\ kg$

Initial velocity of first car is, $u_1=30\vec{j}\ m/s$

Mass of second car is, $m_2=1250\ kg$

Initial velocity of second car is,

Let the combined final velocity after collision be 'v' m/s with as components of final velocity along east and north directions respectively.

Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.

Conserving momentum along the east direction, we have:

Initial momentum = Final momentum

$m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1250\times 39=(1000+1250)v_x\\\\v_x=\frac{48750}{2250}\\\\v_x=\frac{65}{3} m/s$

There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So,

Now, conserving momentum along the north direction, we have:

Initial momentum = Final momentum

$m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\1000\times 30+0=(1000+1250)v_y\\\\v_y=\frac{30000}{2250}\\\\v_y=\frac{40}{3}\ m/s$

There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, $u_{2y}=0$ .

The magnitude of final velocity is given as:

$|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{(\frac{65}{3})^2+(\frac{40}{3})^2}\\\\|\vec{v}|=\sqrt{\frac{5825}{9}}=25.44\ m/s$

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{\frac{40}{3}}{\frac{65}{3}})=31.61^\circ$

So, the final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

7 0

3 years ago