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3 years ago

PLEASE HELP I AM SO STUMPED ABOUT THIS QUESTION PLEASE HELP

Physics

2 answers:

kiruha [24]3 years ago

7 0

This is my alt account just grinding points

Flauer [41]3 years ago

3 0

NO DONT CLICK THE LINK OR COPY IT INTO YOUR BROWSER ITS GONNA PUT A VIRUS ON UR DEVICE

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At what water temperature will additional heat energy need to be added before the temperature will change again?

Lostsunrise [7]

That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)<span />

6 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

Which of the Materials are insulators? (select all that apply)

DENIUS [597]

I believe all the insulators would be glass, wood, plastic, and yarn.

But I’m not entirely sure if mechanical pencil lead is an insulator or conductor.

Hope this helps.

3 0

4 years ago

El conductor de un coche de 650 kg que va a 90 km/h frena y reduce su

Ivanshal [37]

A. 90km/h = 25 m/s
W=1/2mv^2
=1/2* 650* 25^2
= 203125(J)
b. v’= 90-50= 40km/h
= 100/9 m/s
W=1/2mv^2
=1/2*650*(100/9)^2
= 40123,45679

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3 years ago

chanice drives her scooter 7 km north she stops for lunch and then drives 5 km east what distance did she cover. what was her di

liberstina [14]

Use pythagorean's theorem for this, with 7 as a and 5 as b. pythagorean's theorem says that a^2 + b^2 = c^2, so 7^2 * 5^2 = c^2. this gives you 49 + 25 = c^2, so 74 = c^2. c = sqrt 74, which is approximately 8.60 km

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3 years ago