Please help me because i’m trash at these and i need help with five more

How could we expect a white paper to appear when viewed in a red light?

Xelga [282]

The accurate answer is:
You could expect a white paper to have a red tone when viewed in a red light. This is the case because of the neutrality of the white paper.

4 0

3 years ago

14) the formula k2s indicates that

Ivan

14.) c) there are two potassium ions for each ion of sulfur. 15.) b) Lithium fluoride. 16.) d) carbon-12 atom. 17.) <span>c) nonmetal fluorine.</span>

6 0

2 years ago

describe the relationship between the direction of the velocity vector and the direction of the acceleration for a body moving i

Xelga [282]

Answer:

a = v²/r

Explanation:

The acceleration of a body moving in a circular path is known as the centripetal acceleration. This is the acceleration of a body that keeps the body within the circular path. It is written in terms of the linear velocity v and the radius of the circle of rotation as shown;

a = v²/r where

v is the linear velocity

r is the radius

a is the centripetal acceleration

7 0

3 years ago

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .

7 0

2 years ago

PLEASE HELLP ILL GIVE CROWN AND HEART

xxTIMURxx [149]

Answer:

Anti clockwise

hope it helps u

please mark me brailliest

5 0

3 years ago

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