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2 years ago

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What keeps an inflated balloon from falling down if you rub it against your hair and place it against a wall? 1. rubbing distort

s the atoms inside the ballon and polarizes it. 2. rubbing leaves a balloon electrically charged; the charged balloon polarizes the wall. 3. when you rub the balloon against your hair, it will remove some m?

1 answer:

Keith_Richards [23]2 years ago

3 0

2 because when you are doing this it causes friction Which then cause the balloon to stick
~dany-ley

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Suppose you sketch a model of an atom using the ones here as a guide. How would you build a model that is ionized? How would you

kap26 [50]

Answer:

Explanation:

An atom is constructed of three different particles known as electrons, protons and neutrons.

These particles have different mass and charges and are responsible for various characters than an atom posses.

An electron has a negative charge, a proton has positive charge and charge of neutron is neutral. Equal number of electrons and protons are present in an atom that make it electrically neutral but different conditions can occur if we remove these particles from an atom.

1 : Model of an ionized atom - an ionized atom is one which has some net charge on this. It can be either a positive charge or a negative charge.

If we need to sketch the model of an ionized atom then one should either keep the number of electrons less or proton.

2: Model of radioactive atom : A radioactive atom is one an unstable atom and has access of energy in its center. It can be caused by adding either neutrons or protons.

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2 years ago

Bianca is standing at x =600m. Firecracker 1, at the origin, and firecracker 2, at x =900m, explode simultaneously. The flash fr

agasfer [191]

Answer: $3 \mu s$

Explanation:

Given

Bianca is at $x=600 m$

i.e. distance between origin and Bianca is $600 m$

time taken to reach Bianca eyes is

$t=\frac{600}{speed\ of\ light}$

$t=\frac{600}{3\times 10^8}$

$t=2\times 10^{-6} s$

$t=2 \mu s$

i.e. Cracker exploded at $t=2\mu s$ because it is observed at $t=4\mu s$

Time taken by second cracker flash to reach Bianca eyes

$t_2=\frac{300}{3\times 10^8}$

$t_2=10^{-6}$

$t_2=1 \mu s$

Therefore it will be observed at $t=3 \mu s$

4 0

2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

If you have an effusive personality, then it __________ of you. motor oil will __________ across the gravel if it spills out of

erica [24]

The first answer in the blank is pours out of you, the second answer in the space provided is diffuse. It is because a person who has an effusive personality will likely pour out his or her emotions while the motor oil will likely diffuse if this has been spilled out of the can.

6 0

2 years ago

c) If the ice block (no penguins) is pressed down even with the surface and then released, it will bounce up and down, until fri

eduard

Answer:

y = 20.99 V / A

there is no friction y = 20.99 h

Explanation:

Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy

when the block of ice is submerged it is subjected to two forces its weight hydrostatic thrust

F_net= ∑F = B-W

the expression stop pushing is

B = ρ_water g V_ice

where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice

We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³

W = ρ-ice g V

F_net = (ρ_water - ρ_ ice) g V

this is the net force directed upwards, we can find the potential energy with the expression

F = -dU / dy

ΔU = - ∫ F dy

ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy

ΔU = - (ρ_water - ρ_ ice) g A y² / 2

we evaluate between the limits y = 0, U = 0, that is, the potential energy is zero at the surface

U_ice = (ρ_water - ρ_ ice) g A y² / 2

now we can use the conservation of mechanical energy

starting point. Ice depth point

Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2

final point. Highest point of the block

$Em_{f}$ = U = m g y

as there is no friction, energy is conserved

Em₀ = Em_{f}

(ρ_water - ρ_ ice) g A y² / 2 = mg y

let's write the weight of the block as a function of its density

ρ_ice = m / V

m = ρ_ice V

we substitute

(ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y

y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A

let's substitute the values

y = 0.913 / (1 - 0.913) 2 V / A

y = 20.99 V / A

This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction

V / A = h

where h is the height of the block

y = 20.99 h

7 0

2 years ago