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3 years ago

14

I need help one through five

1 answer:

Mama L [17]3 years ago

7 0

Thinking critically is basically thinking outside of the box, so it's asking how does the bulub filament change after it's on, a filament is basically the wiring inside.

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The terms sanity and insanity:

IgorLugansk [536]

Whats the question exactly?

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2 years ago

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How many feet are in a mile

klemol [59]

Hello There!

There are 5,280 feet in 1 mile.

Have A Great Day!

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3 years ago

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What are some ways in which body fat can be measured

MissTica

I only know about the Water tank which is the most accurate. You place your body in it, and weights are added I think. Somehow some measurements are gathered to get your body fat weight. Online calculators exist, as well as electronic waves that are sent int your body, the echo is recorded and analyzed.

8 0

3 years ago

A box slides down a ramp inclined at 24◦ to the horizontal with an acceleration of 1.7 m/s 2 . The acceleration of gravity is 9.

dsp73

Answer:

<h3>0.445</h3>

Explanation:

In friction, the coefficient of friction formula is expressed as;

$\mu = \frac{F_f}{R}$

Ff is the frictional force = Wsinθ

R is the reaction = Wcosθ

Substitute inti the equation;

$\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta$

Given

θ = 24°

$\mu = tan 24^0\\\mu = 0.445\\$

Hence the coefficient of kinetic friction between the box and the ramp is 0.445

3 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago