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3 years ago

Anybody wanna play among us?

Physics

2 answers:

kherson [118]3 years ago

8 0

Answer:

yess

Explanation:

Grace [21]3 years ago

3 0

Answer:

yes me

<h3>mark me brainliest</h3>

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What would the kinetic energy and potential energy be with the snowboarder, before during and after a jump?

Reika [66]

Answer: Before the jump, the snowboarder would carry potential energy.

During the jump he will carry kinetic energy.

And after the jump, assuming hes at a full stop, he will carry potential energy once again.

3 0

3 years ago

The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm

lord [1]

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: $n_0$ = 1.52

Wave length of light = λ = 570 nm = $570\times10^{-9}\ m$

$\text{ Thickness}=\dfrac{\lambda}{4n}$

$=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}$

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

5 0

3 years ago

Please help im being timed on this and need to get my grade up

nadya68 [22]

Answer:

a I think hope this helps

8 0

2 years ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$