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3 years ago

The light you see around you from the Sun is called UVvisibleX-infrared rays.

1 answer:

6 0

The correct answer for the question that is being presented above is this one: "Ultraviolet rays." The light you see around you from the Sun is called ultraviolet rays. Sunlight is a portion of the electromagnetic radiation given off by the Sun, in particular infrared, visible, and ultraviolet light.

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Adding a best-fit line to the scatter plot shown below would be an example of _____.

baherus [9]

I think It would be C. Checking a prediction. Sorry if I’m wrong

3 0

3 years ago

Read 2 more answers

A ball is kicked at a speed of 16m/s at 33° and it eventually returns to ground level further down field.

saw5 [17]

Hi there!

We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).

Remember to break the velocity into its vertical and horizontal components.

Thus:

0 = vi - at

0 = 16sin(33°) - 9.8(t)

9.8t = 16sin(33°)

t = .889 sec

Find the max height by plugging this time into the equation:

Δd = vit + 1/2at²

Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²

Solve:

Δd = 7.747 - 3.873 = 3.8744 m

4 0

3 years ago

Read 2 more answers

No links please, links never work, this is science not physics

stiks02 [169]

Answer:

b and c are the answers

A is an opinion, D is a superstitious belief because they haven’t found Jesus obviously

7 0

3 years ago

Can you answer the b. thanks

levacccp [35]

Answer:

current going into a junction in a circuit is EQUAL TO the current comming out of the junction.

Explanation:

Krichhoff's Current Law

Kirchhoff's current law (1st Law) states that current flowing into a node (or a junction) must be equal to current flowing out of it.

4 0

3 years ago

A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol

PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a) at 4.75 cm

b) at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula $E = \dfrac{kq }{d}$

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}$

E = 1461.95 N/C

c) The electric field E is calculated as:

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}$

E = 239.76 N/C

7 0

3 years ago