In order for a color to be seen by a human, the light must be what?

Answers

Answer 1

Answer:
The light must be within a certain, rather narrow, range
of wavelength/frequency, and it must enter his eye.

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How can you cause a physical change?

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Take the sample and . . .
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Group__ contains elements like fluorine chlorine bromine iodine and astatine

Answers

Group 17 or vii contain element like fluorine, chlorine, bromine, iodine and astatine

explanation

Group 17 element in periodic table are referred to as halogen.
They have 7 valence electron
They are highly reactive
Have a low melting and boiling point but increases down the group
They have a tendency to form salt
There electronegativity decreases down the group.

Where there any mammal dinosaurs

Answers

I believe that the first mammals came after the dinosaurs died

No they were absolutely none of them died off before the first memo was born

An airplane is moving at 350 km/hr. If a bomb isdropped from the airplane at 1.5 km, (a) with what
velocity does the bomb strike the earth? (b) How
long does it take the bomb to fall? (c) What is the
bomb's range?

Answers

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_(o)+V_(oy)t-(1)/(2)gt^(2)$ (1)

$x=V_(ox)t$ (2)

$V_(f)=V_(oy)-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_(o)=1.5 km (1000 m)/(1 km)=1500 m$ is the bomb'e initial height

$V_(oy)=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^(2)$ is the acceleration due gravity

$x$ is the bomb's range

$V_(ox)=350 (km)/(h) (1000 m)/(1 km) (1 h)/(3600 s)=97.22 m/s$ is the bomb's initial horizontal velocity

$V_(f)$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

b) Time

With the conditions given above, equation (1) is now written as:

$y_(o)=(1)/(2)gt^(2)$ (4)

Isolating $t$ :

$t=\sqrt{(2 y_(o))/(g)}$ (5)

$t=\sqrt{(2 (1500 m))/(9.8 m/s^(2))}$ (6)

$t=17.49 s$ (7)

a) Final velocity

Since $V_(oy)=0 m/s$ , equation (3) is written as:

$V_(f)=-gt$ (8)

$V_(f)=-(97.22)(17.49 s)$ (9)

$V_(f)=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

c) Range

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

Final answer:

The bomb hits the earth at approximately 540.3 km/hr, it takes approximately 17.5 seconds for the bomb to hit the ground, and the bomb's range is approximately 1.36km.

Explanation:

To answer these questions, we're going to use the principles of kinematics that involve motion in a straight line with constant acceleration (that is the acceleration due to gravity in this case). (a) The velocity of the bomb when it strikes the earth can be calculated using the kinematic equation v = u + gt, where u is the initial velocity (0 m/s since the bomb was dropped, not thrown), g is the acceleration due to gravity (approximately -9.8 m/s²), and t is the time. However, we do not know the time yet, so we will use the equation v = sqrt(2×g×h), where h is the height (1500m), and this gives us approximately 540.3 km/hr. (b) The time it takes for the bomb to fall can be calculated using t = sqrt((2×h)/g) which gives approximately 17.5 seconds. (c) The range of the bomb is the horizontal distance it travels, which is the velocity of the plane multiplied by the time it takes for the bomb to hit the ground, or 350 km/hr × 17.5/3600 hr = approximately 1.36km.

Learn more about Kinematics here:

brainly.com/question/7590442

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The height of a flagpole is 8.5 meters tall. If the length of the meter stick shadow is 1.5 meters long ,what would be the the length of the flagpoles shadow?

Answers

Explanation:

Similar triangles:

1.5 m / 1 m = x / 8.5 m

x = 12.75 m

Round as needed.

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