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3 years ago

Who is this and what show does she play in?

Physics

1 answer:

ryzh [129]3 years ago

7 0

Answer:

bella thorne

Explanation:

i think that is her, and she is dyslexic!

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Gordon throws a baseball into the air. It rises, stops when it reaches its greatest height, and then falls back to the ground. A

dangina [55]

Both C and D are technically correct. But I’d go with D in this instance.

5 0

3 years ago

At what angle should the axes of two polaroids be placed so as to reduce the intensity of the incident unpolarized light to 13.

prohojiy [21]

Answer:

θ = 66.90°

Explanation:

we know that

$I= \frac{I_0}{2}cos^2\theta$

I= intensity of polarized light =1

I_o= intensity of unpolarized light = 13

putting vales we get

$1= \frac{13}{2}cos^2\theta$

⇒ $\theta=cos^{-1} \sqrt{\frac{1}{6.5} }$

therefore θ = 66.90°

5 0

3 years ago

When sound waves are reflected back to the source, the reflected sound is called

sergey [27]

It’s an echo :))))))

6 0

3 years ago

Read 2 more answers

An antibaryon composed of two antiup quarks

Sveta_85 [38]

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have $+\frac{2}{3} e$ charge where as down, strange and bottom quarks have $-\frac{1}{3}e$ charge.

The antiparticle of up quark is antiup quark and has charge $-\frac{2}{3}e$ charge.

The antiparticle of down quark is antidown quark and has charge $+\frac{1}{3}e$ charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

$2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e$

Thus, antibaryon has -1e charge.

5 0

4 years ago

1) A particle move along the x axis, as x=tsquare-2t+1(SI unit). (a) The functions of its velocity and accelerator vector agains

Bond [772]

Answer:

$\text{I believe this is calculus, no? But I believe that there should be a y-component of position}\\\text{But based off of the information provided, here goes:}\\\text{(a)}\\x'(t)=v(t)=\frac{d}{dt}t^2-2t+1\\v(t)=2t-2\text{ is the velocity vector}\\v'(t)=a(t)=\frac{d}{dt}2t-2\\a(t)=2 \text{ is the acceleration vector}\\\text{(b)}\\\text{Average Velocity }=\frac{x_{f}-x{i}}{t_{f}-t_{i}}\\=\frac{1-1}{2}=0, \text{based off of the information provided, the average velocity in the first 2 seconds}: 0m/s$

8 0

3 years ago