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2 years ago

Which of newton's laws explains why your hands get red when you press them hard against a wall

1 answer:

5 0

The Newton’s law that explains why the hands get red when you press them hard against a wall is Newton’s third law. When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite direction on the first body.

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Which of the following is true of education in 1950

Maksim231197 [3]

Information I learned from history class Education in the 1950's expanded from previous decades. They no longer focused purely on reading, writing and arithmetic. History and science became a main part of the cirriculum. Also, enrollment skyrocketed as the baby-boomers began enrolling in elementary school. One interesting thing that categorized this generation was the presence of fallout tests. Schools would require the students to go through a fake atomic bomb attack in which they would hide under their desks (which was completely pointless in protecting them from radiation, it was more of an emotional security for the parents and teachers, but scared the hell out of the students). Socially, children were taught to conform and to be normal. Standing out or questioning authority was bad. Sex was taught, though minimally. They explained the penis and vagina. Sexually transmitted diseases were focused on greatly so as to "scare" the students out of premarital sex.

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3 years ago

Can you help me doing an essay about actual self

Rudik [331]

Answer:

yes absolutely

Explanation:

why not

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2 years ago

If 20 beats are produced within a single second, which of the following frequencies could possibly be held by two sound waves tr

zhuklara [117]

The correct choice is

D. 22 Hz and 42 Hz.

In fact, the beat frequency is given by the difference between the frequencies of the two waves:

$f_B = |f_1 -f_2|$

In this problem, the beat frequency is $f_B=20 Hz$ , therefore the only pair of frequencies that gives a difference equal to 20 Hz is

D. 22 Hz and 42 Hz.

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2 years ago

Read 2 more answers

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:

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

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:

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