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3 years ago

9

Is the color spectrum simply a small segment of the electromagnetic spectrum?

1 answer:

ch4aika [34]3 years ago

3 0

Answer:

yup, u r correct

Explanation:

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Hahahahaha how do you give ppl brainliest

sergeinik [125]

Answer:

When you ask a question, only two people can answer. When there are two answers, a little crown should appear at the bottom right hand corner. All you have to do is click the crown and it gives Brainliest. But you can only give it to one person per question

Explanation:

8 0

3 years ago

What is the difference between a chemical change and a physical change? *

slavikrds [6]

Answer:

I do belive that it is B hrs cn I an gn

7 0

3 years ago

A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo

Svetllana [295]

Answer:

The gravitational potential energy of the nickel at the top of the monument is 8.29 J.

Explanation:

We can find the gravitational potential energy using the following formula.

$GPE=mgh$

Identifying given information.

The nickel has a mass $m=0.005 \,kg$ , and it is a the top of Washington Monument.

The Washington Monument has a height of $h=555 \, ft$ , thus we need to find the equivalence in meters using unit conversion in order to find the gravitational potential energy.

Converting from feet to meters.

Using the conversion factor 1 m = 3.28 ft, we have

$h = 555 \, ft \times \cfrac{1 \, m}{3.28 \, ft}$

That give u s

$h = 169.2 \, m$

Finding Gravitational Potential Energy.

We can replace the height and mass on the formula

$GPE=mgh$

And we get

$GPE=(0.005)(9.8)(169.2) \, J$

$\boxed{GPE=8.29 \,J}$

The gravitational potential energy of the nickel at the top of the monument is 8.29 J.

7 0

3 years ago

Read 2 more answers

Are moons 1-4 waxing are waning ?

solong [7]

Answer:

I want to help you but i cant.

Explanation:

please provide a screenshot or photos of moons 1-4

3 0

2 years ago

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

goldfiish [28.3K]

Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

$y=y_{0}+uSin\theta t-1/2 gt^{2}$

$y=2+20Sin5 t-4.9t^{2}$

As it hits the ground in time t, so put y = 0

$0=2+1.74 t-4.9t^{2}$

$4.9t^{2}-1.74t-2=0$

$t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}$

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X = 16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

$y'=y_{0}+uSin\theta t'-1/2 gt'^{2}$

$y'=2+20Sin5 t-4.9\times0.35^{2}$

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

3 0

2 years ago