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3 years ago

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When you rub a glass rod with silk. The rod becomes positive charged, and the silk becomes negative.Also when you rub a rubber r

od with fur, the rubber rod becomes negative and the fur becomes positively charge.Question:1. What happens if you rub a glass rod with fur?2. What happens if you rub a rubber rod with silk?

1 answer:

adelina 88 [10]3 years ago

7 0

Answer:

Not sure

Explanation:

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Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

<u></u>

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 12 blades and rotates at a

tester [92]

Answer:

v<em>min</em> = 0.23 m/s

Explanation:

The golf ball must travel a distance equal to its diameter in the time between blade arrivals to avoid being hit. If there are 12 blades and 12 blade openings and they have the same width, then each blade or opening is 1/24 of a circle of is 2π/24 = 0.26 radians across.

Therefore, the time between the edge of one blade moving out of the way and the next blade moving in the way is

time = angular distance/angular velocity

⇒ t = 0.26 rad / 1.35 rad/s = 0.194 s

The golf ball must get completely through the blade path in this time, so must move a distance equal to its diameter in 0.194 s, therefore the speed of the golf ball is

v =d/t

⇒ v = 0.045 m / 0.194 s = 0.23 m/s

7 0

3 years ago

The __________ on the left side of an equation must be balanced with the atoms on the right side of the equation.

zhannawk [14.2K]

Your answer would be total number of atoms! This is because when you have these equations which require total number of atoms.

3 0

3 years ago

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During Mr. Nye's science class, students were expected to identify various substances using physical properties they could easil

nydimaria [60]

Density is the best property to use, as while multiple different metals could create cubes with the same color, mass, or volume, no different metal could create a cube with the same mass and volume. Density is based on mass and volume, and as a result no two different metals will have the same density.

4 0

3 years ago

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A 110-g bullet is fired from a rifle having a barrel 0.636 m long. Choose the origin to be at the location where the bullet begi

astraxan [27]

<span>Data:
mass = 110-g bullet
d = 0.636 m
Force = 13500 + 11000x - 25750x^2, newtons.

a) Work, W

W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =

W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636 =

W = 8602.6 joule

b) x= 1.02 m

</span><span><span>W = 13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02

W = 10383.5

c) %

[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.

</span>

7 0

3 years ago