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2 years ago

For an object to appear red, what must happen?

Physics

1 answer:

Lapatulllka [165]2 years ago

3 0

Answer:

<h2>C. It must reflect red light wavel</h2>

Explanation:

The color of the object is reflected by the object other colors are absorbed by the object.

Here object appears in red color so red color is reflected by the object and other colors are absorbed.

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9- Under what circumstances would a vector have components that are equal in

valkas [14]

Explanation:

c. if the vector is oriented at 0° from the X -axis.

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2 years ago

Compare and contrast offensive and defensive roles in team sports.

strojnjashka [21]

In the offensive role, the players try to get a goal.
In the defensive roll, The players try to protect the goal
Hoped this helped a little :)

4 0

2 years ago

Read 2 more answers

A planet with a mass one-half that of Earth has a radius that is 3 times that of Earth's radius. What is the gravitational field

stiv31 [10]

The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.

So when you stand on the surface of this particular planet, you feel a force of gravity that is

(1/2) / (3²)

of the force that you feel on the surface of the Earth.

That's (1/18) as much as on Earth.

The acceleration of gravity there would be about 0.545 m/s².

This is about 12% less than the gravity on Pluto.

6 0

2 years ago

Acceleration due to gravity is stated as a constant which is given 9.8 m/s2. Many scientists have planned and design experiments

stich3 [128]

The answer is 98.1 m/s2.

3 0

2 years ago

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

3 0

2 years ago