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3 years ago

What effect does observing a substance s physical properties have on the substance

Physics

1 answer:

Aneli [31]3 years ago

4 0

It can be observed without changing the identity of the object.

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The Cartesian coordinate of a point in the xy plane are (x,y)=(-3.50,-2.50)m. Find the poler coordinate of this point

Masja [62]

Answer:

The polar coordinate of $P(x,y) = (-3.50\,m,-2.50\,m)$ is $P (r,\theta) = (4.301\,m, 215.538^{\circ})$ .

Explanation:

Given a point in rectangular form, that is $P(x,y) = (x,y)$ , its polar form is defined by:

$P(x,y) = (r,\theta)$ (1)

Where:

$r$ - Norm, measured in meters.

$\theta$ - Direction, measured in sexagesimal degrees.

The norm of the point is determined by Pythagorean Theorem:

$r = \sqrt{x^{2}+y^{2}}$ (2)

And direction is calculated by following trigonometric relation:

$\theta = \tan^{-1} \frac{y}{x}$ (3)

If we know that $x = -3.50\,m$ and $y = -2.50\,m$ , then the components of coordinates in polar form is:

$r = \sqrt{(-3.50\,m)^{2}+(-2.50\,m)^{2}}$

$r \approx 4.301\,m$

Since $x < 0\,m$ and $y < 0\,m$ , direction is located at 3rd Quadrant. Given that tangent function has a period of 180º, we find direction by using this formula:

$\theta = 180^{\circ}+\tan^{-1} \left(\frac{-2.50\,m}{-3.50\,m} \right)$

$\theta \approx 215.538^{\circ}$

The polar coordinate of $P(x,y) = (-3.50\,m,-2.50\,m)$ is $P (r,\theta) = (4.301\,m, 215.538^{\circ})$ .

5 0

2 years ago

An object that is magnetic all of the time is called a.

Novosadov [1.4K]

Answer:

Bar magnets are permanent magnets. This means that their magnetism is there all the time and cannot be turned on or off as it can with electromagnets .

Explanation:

5 0

2 years ago

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) what fraction of its initial energy is los

Scilla [17]

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100

and since mass and gravity are constant so it leaves us with just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the energy or 0.25 in fraction

b) Here use the equation vf^2=vi^2+2gd

where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable
we know that at maximum height or peak, the velocity is 0 so vf is 0

therefore,

vi =sqrt(-2gd) 
vi =sqrt(-2x-9.81x1.5) 
vi =5.4 m/s

c) The energy was converted to heat due to friction with the air and the ground.

6 0

3 years ago

In which situations is heat flow present? Check all that apply.

babunello [35]

A substance feels warm

7 0

3 years ago

Read 2 more answers

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,

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