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1 year ago

Which statement best explains why an object appears green in sunlight?

Physics

2 answers:

Lesechka [4]1 year ago

5 0

Answer:

Explanation:

More green light is reflected while more red and blue light is absorbed.

Helen [10]1 year ago

3 0

Answer:

This question is incomplete, the options are:

A) The object absorbs most white light and refracts most green light.

B) The object refracts most white light and absorbs most green light.

C) More green light is absorbed while more red and blue light is reflected.

D) More green light is reflected while more red and blue light is absorbed.

The answer is D.

Explanation:

Light is an electromagnetic wave that contains different colours at different wavelength. The colour of light that is seen depends on the wavelength of light that is REFLECTED, while other wavelengths of light are ABSORBED. This feature is dependent on the properties of each object that received the sunlight.

For example, an object will appear GREEN because it has properties that enables it to REFLECT most of the GREEN LIGHT but absorbs most of the RED AND BLUE LIGHT in the sunlight passing through the object.

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A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app

Cerrena [4.2K]

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

6 0

1 year ago

Question 2 (1 point)

love history [14]

Answer:

neutral

Explanation:

3p - 3e = 0 and that leaves 2 neutrons so it will be neutral

6 0

2 years ago

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Football player A has a mass of 210 pounds and is running at a rate of 5.0 mi/hr. He collides with player B. Player B has a mass

Naddika [18.5K]

The equation for momentum is p = mv where p is the omentum, m is the mass and v is the velocity. Calculating the momentum for each football player, player A will have a momentum of 1050 lb-mi/h and player B will have a momentum of 570 lb-mi/h. Therefore, momentum of player A is greater than that of player B.

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1 year ago

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Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acce

Lana71 [14]

Answer:

Explanation:

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acceleration of the two blocks be equal to zero

F - Force

T = Tension

m = mass

a = acceleration

g = gravitational force

Let the given Normal on block 2 = N

and $N = m_2 g \cos \theta$

and the tension in the given string is said to be $T = m_2 g \sin \theta$

When the acceleration $a=\frac{F}{m_1}$

for the said block 1.

It will definite be zero only when Force is zero , F=0.

Here by Force, F

I refer net force on block 1.

Now we know

$F = m_1g-T.$

It is known that if the said

$\theta=\frac{\pi}{2}$ ,

then Tension $T= m_2g$ $[since \sin(\pi/2) = 1]$ ,

Now making $"F = m_1g - m_2g"$

So If we are to make Force equal to zero

$F=0 => m_1g = m_2g \ or \ m_1 = m_2$

6 0

1 year ago

A robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free-fall acceleration is −3.7 m/s2 .

ira [324]

<span>a. We can find the velocity when the camera hits the ground. v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera hits the ground with a velocity of 42 m/s b. We can find the time it takes for the camera to hit the ground. y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds

It takes 11.4 seconds for the camera to hit the ground.</span>

3 0

1 year ago