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3 years ago

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Standing on a mountain, you look towards another mountain 6.00 km away, on which there are two persons standing 1.00 meter apart

. If the pupil of your eye has a diameter of 3.50 mm, will you be able to resolve the two persons (or will you see them as one). Take wavelength of light = 600 nm

1 answer:

allsm [11]3 years ago

5 0

Answer:

No.

Explanation:

We shall solve this problem by calculating the resolving power of eye for given wavelength

Resolving Power of eye = \frac{1.22\lambda }{D}

Where λ is wave length of light and D is diameter of eye.

λ is 600 nm and D is 3.5 mm . Put these values in the given formula

Resolving Power = \frac{1.22\times 600\times 10^{-9} }{3.5\times 10^{-3}}\\

=209.14 \times 10^{-6}radian

From the formula

Φ = \frac{L}{D}[/tex]

Where Ф is resolving power . If L be distance between two points that can be resolved at distance D. D is 6 km or 6000 m .

209.14 \times 10^{-6}=\frac{L}{6000}\\

L= 1.254 m

So minimum distance that can be resolved is 1.254 m.

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A 23.3 kg crate is pushed with a force of 944 N for 36.0 seconds, moving the crate a distance of 12.4 m. How much power was used

tigry1 [53]

Explanation:

$p = \frac{f \times s}{t}$

power = Force × distance /time

power = 944N × 12.4m/36secs

power = (944×12.4/36)Nms—¹

power = 390.2Nms—¹ or 390.2Watts or 390.2Js—¹

8 0

3 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

Fill in the blanks HELP ASAP

Alexandra [31]

checking to see if im locked out

7 0

3 years ago

Please help me ..im begging you

Deffense [45]

Explanation:

let's assume that:

v1= 600ml=0,6l

T1=27°C= 300K

p1=700mmHG=93326Pa

T2=-20°C=253K

p2=500mmHg=66661

V2=?

p1V1/T1=p2V2/T2 => V2=p1V1T2/p2

V2= 93326*0,6*253/66661

V2=212,52l

5 0

3 years ago

A 41.0 kg child swings in a swing supported by two chains, each 2.98 m long. (a) If the tension in each chain at the lowest poin

Alex777 [14]

Answer:695.5 N

Explanation:

mass of child $m=41 kg$

Length of chain $L=2.98 m$

Tension in each chain $T=348 N$

(a)Tension at bottom point $T=348 N$

At lowest Point

$T+T-mg=\frac{mv^2}{L}$

$2T-mg=\frac{mv^2}{L}$

$2\times 348-41\times 9.8=\frac{41\times v^2}{2.98}$

$v^2=\frac{294.2\times 2.98}{41}$

$v=\sqrt{21.38}=4.62 m/s$

(b)Force exerted by Seat will be Equal to Normal reaction

$N-mg=\frac{mv^2}{L}$

$N=mg+\frac{mv^2}{L}$

$N=41\times 9.8+\frac{41\times 21.38}{2.98}$

$N=695.95 N$

8 0

3 years ago