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3 years ago

10

An inflated balloon was pressed against a wall after it has been rubbed with a piece of synthetic cloth. It was found that the b

alloon sticks to the wall. what force might be responsible for the attraction between the balloon and the wall ????

2 answers:

Anna71 [15]3 years ago

6 0

Answer: friction

Explanation:

Because when you rub to objects together it creates static electricity

serg [7]3 years ago

3 0

Friction! Because the ballon was rub on the cloth in order to stick to the wall

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A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

b

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

c

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

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A rock falls from a cliff and hits the ground at a velocity of 31m/s. How long did it take to fall?

Paladinen [302]

Answer:

i dont know man

Explanation:

i dont know man

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The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th

pickupchik [31]

Answer:

a

The focal length of the lens in water is $f_{water} = 262.68 cm$

b

The focal length of the mirror in water is $f =79.0cm$

Explanation:

From the question we are told that

The index of refraction of the lens material = $n_2$

The index of refraction of the medium surrounding the lens = $n_1$

The lens maker's formula is mathematically represented as

$\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

Where $f$ is the focal length

$n$ is the index of refraction

$R_1 and R_2$ are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air we have

$\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

When immersed in liquid the formula becomes

$\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ]$

The ratio of the focal length of the the two medium is mathematically evaluated as

$\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }$

From the question

$f_{air }$ = 79.0 cm

$n_2 = 1.55$

and the refractive index of water(material surrounding the lens) has a constant value of $n_1 = 1.33$

$\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}$

$f_{water} = 262.68 cm$

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

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When a wave slows down it...

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<span>a. bends away from the normal</span>

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A convex mirror curves ___________.<br><br> what does a convex mirror curves?

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So, a concave mirror is a mirror that goes inward, thus, making a convex mirror have an outward curvature. Ex: A convex mirror is used at the front of a bus.

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