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3 years ago

Which statement describes what happens when a balloon is rubbed with a wool cloth?

2 answers:

8 0

Electrons move from the atoms in the cloth to the atoms in the balloon, causing the balloon to have a negative charge

Hope this helps!!!

sladkih [1.3K]3 years ago

4 0

Yes, I just took this test, and can 100% confirm the correct answer to this question is e<span>lectrons move from the atoms in the cloth to the atoms in the balloon, causing the balloon to have a negative charge.

Hope this helps, and have a great day! :)</span>

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A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A

Kruka [31]

Answer:

$\omega_f = 585.37 \ rev/s$

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)= 800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

$I \omega_i = I' \omega_f$

$(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f$

$M\times 800 = ( M + m )\omega_f$

$3\times 800 = (3+1.1)\times \omega_f$

$2400 = (4.1)\times \omega_f$

$\omega_f = 585.37 \ rev/s$

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4 years ago

In the following equation , which compounds are the reactants? NaCL+AgNo>NaNo+AgCl

NeX [460]

The first two are always the reactants the products come after so they are last

3 0

3 years ago

Water flowing through a small creek into a stream after a storm is an example of _____.

Serggg [28]

Runoff water i belive

7 0

3 years ago

The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

Liula [17]

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

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4 years ago

9. A balloon is filled with 1000 cm3 of a gas weighing 1000 g. Will it rise or fall when it is released?

torisob [31]

We can calculate the density of the balloon as follows:

$\rho=\frac{mass}{volume}=\frac{1000g}{1000cm^3}=\frac{1g}{cm^3}$

Therefore, the balloon will fall

Since the density of air is about 0.00123 g/cm^3 , the balloon is much more dense than the surrounding air. As a result, the balloon weighs more than the air that it displaces so the balloon will fall.

3 0

1 year ago