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3 years ago

6

A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffract

ion standpoint is

1 answer:

Anestetic [448]3 years ago

8 0

In the near ultraviolet

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A toy car is placed at 0 on a number line. It moves 9 cm to the left, then 4 cm to the right, and then 6 cm to the len

MA_775_DIABLO [31]

Answer:19 cm

Explanation:

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2 years ago

Read 2 more answers

How does gravity keep the moon in orbit around earth

lidiya [134]

Answer:

The moon's orbit draws the oceans to it, which triggers ocean tides. Force produces stars and planets by gathering the mass from which it exists.

Explanation:

The moon's orbit draws the oceans to it, which triggers ocean tides. Force produces stars and planets by gathering the mass from which it exists.

Answer is above

<em><u>Hope this helps.</u></em>

3 0

2 years ago

How to solve 30(a)<br><br> Give solution asap.

expeople1 [14]

Answer:

They will not meet

h-hX=1.2*g*t²

hX=v0*t-(1/2*g*t²)

Explanation:

fall h=1/2*g*t²

elevation time if v0=20 m/s te=v0/g=20 m/s /9.81 m/s²=2.0387s

hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m

free fall

t=2.0387s yields hX=1/2*g*t²=20.387 m

h-hX=200m - 20.387 m=179,613 m.

so, the second body has not enough initianoal speed to reach a meeting point

5 0

3 years ago

Another word for stored energy is <br> 1 potential <br> 2 kinetic

scoray [572]

Answer:

Potential

Explanation:

Potential energy is the energy stored while kinetic energy is motional energy.

Potential itself means "having capacity/energy".

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3>

8 0

2 years ago

The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =

dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar, θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

∑ Mₐ = Iα

Where, Mₐ - the moment about A

α - angular acceleration

I - moment of inertia of the rod AB

$-mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha$

$\alpha=\frac{-3gcos\theta}{2l}$

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

$\vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

The acceleration at the center of the bar

$\vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}$

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]$

Comparing the coefficients of i

$=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}$

Comparing coefficients of j

$(\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}$

Net force on x direction

$F_{x}=(\vec{a_{G}})_{x}$

substituting the values

$F_{x}$ =1.5(14.58L+11.96)

Similarly net force on y direction

$F_{y}=(\vec{a_{G}})_{y}+mg$

= 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}$

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

8 0

3 years ago