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3 years ago

13

Electrically inert metal ball A is connected to the ground by a wire. What happens to the charge of this ball if you bring a neg

atively charged ball B near it. Justify your answer.

1 answer:

kaheart [24]3 years ago

7 0

Explanation:

They will repel, meaning that they are made of an electrical conductor.

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Which of the following are correct statements about the way an atom is put

Ann [662]

Answer:

valenc e shell

Explanation:

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3 years ago

A 68 kg object, starting from rest, travels from point A to point B at a rate of 30 m/s in 2 hours. What is the applied force on

Natasha2012 [34]

Answer:

$\huge\boxed{\sf F = 0.28\ N}$

Explanation:

<h3>Given Data:</h3>

Mass = m = 68 kg

Velocity = v = 30 m/s

Time = 2 hours = 2 × 60 × 60 = 7200 s

<h3>Required:</h3>

Force = F = ?

<h3>Formula to be used:</h3>

$\displaystyle F = \frac{mv}{t}$

<h3>Solution:</h3>

$\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}$

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1 year ago

What type of telescope is shown in Figure 24-2

lesantik [10]

Refractor, It's a refractor-esque telescope

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3 years ago

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An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

a)

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

b)

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

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3 years ago

How can parents help children to gain friends?

Andre45 [30]

Answer:

You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.

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3 years ago

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