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4 years ago

If a substance cannot be separated by physical means or chemical means, it is ___________.

Physics

2 answers:

Aleks04 [339]4 years ago

8 0

Answer:

If a substance cannot be separated by physical means or chemical means, it is an element.

Explanation:

A chemical element is a substance that cannot be broken down into other substances by chemical means and methods. An element consists of one kind of atoms with all the same atomic number, so with the same number of protons in the nucleus. The number of neutrons in the atoms of an element can vary.

The elements are subdivided into metals, metalloids and non-metals. Most elements are a solid at room temperature. Exceptions include the noble gases that are gaseous, and mercury, which is liquid.

ELEN [110]4 years ago

4 0

If a substance cannot be separated by physical means or chemical means, it is an element. Elements<span> are chemically the simplest substances and hence cannot be broken down using </span>chemical or physical<span> methods. Hope this answers the question. Have a nice day.</span>

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bearhunter [10]

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3 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

7 0

4 years ago

What is Newton's Law

r-ruslan [8.4K]

I hope you are referring to Newton's Laws of Motion by "Newton's Law".

There are three laws quoted by the great physicist of all time Sir Isaac Newton.

These laws are the building-blocks of the field of the Physics known as Classical Physics or Classical Mechanics.

Law 1. If no force applied externally, then a body in rest will always be in rest and a body in motion will continue to move in a straight line with a uniform velocity.

Law 2. The rate of change of momentum is directly proportional to the force applied and the direction of motion is always in the direction of the force applied.

Law 3. Every action has an equal and opposite reaction.

The Law 1 is also called the law of Inertia.

From the 2nd law, we can derive the equation of force, that is F =m.a (m = mass; a = acceleration)

5 0

4 years ago

A ball is tossed vertically upwards with a speed of 5.0 m s–1. After how many seconds will the ball return to its initial positi

mezya [45]

Answer:

1.02 seconds

Explanation:

Initial velocity = u = 5 m/s

t = Time taken

s = Displacement travelled = 0

a = Acceleration due to gravity = -9.81 m/s² (negative sign due to direction)

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow 0=5t+\frac{1}{2}-9.81\times t^2\\\Rightarrow 0=5t-4.905t^2\\\Rightarrow 0=t(5-4.905t)\\\Rightarro 0=5-4.905t\\\Rightarrow t=\frac{-5}{-4.905}=1.02\ seconds$

So, time taken to return to its initial position is 1.02 seconds

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4 years ago

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3 years ago