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4 years ago

Please help! I especially need help with the second question but help with the first one would be most appreciated!

Physics

1 answer:

lara31 [8.8K]4 years ago

4 0

Answer:

a) Team A will win.

b) The losing team will accelerate towards the middle line with 0.01 m/ $s^{2}$

Explanation:

Given that Team-A pulls with a force , $F_{1} = 50N$

and Team-B pulls with a force , $F_{1} = 45N$

∵ $F_{1} > F_{2}$

The rope will move in the direction of force $F_{1}$ .

∴ Team-A will win.

b) Considering both the teams as one system of total mass , $m = 246+253 = 499 kg$

Net force on the system , $F = F_{1} - F_{2}$ = 50-45 = 5N

Applying Newtons first law to the system ,

F = ma , where 'a' is the acceleration of the system.

Since , both the teams are connected by the same rope , their acceleration would be the same.

∴ 5 = 499×a

∴ a = 0.01 m/ $s^{2}$

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3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

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A. 6.67 cm

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$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

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$s'=-Ms$

and then we can substitute it into the lens equation:

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and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

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Now the image distance can be directly found by using again the magnification equation:

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And the sign of s' (negative) also tells us that the image is virtual.

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3 years ago

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where,

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