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Vera_Pavlovna [14]
3 years ago
6

Heptane is always composed of 84.0% carbon and 16.0% hydrogen. This illustrates the law of multiple proportions. conservation of

mass. definite proportions. all of the above
Physics
1 answer:
BigorU [14]3 years ago
5 0

Heptane is always composed of 84.0% carbon and 16.0% hydrogen. This illustrates the "law of definite proportions".

Answer: Option C

<u>Explanation:</u>

Proust's law states that every chemical compound used to made up of element constituents with constant proportions in terms of its mass and also independent from its sources and synthesis method. In 1779, Joseph Proust gave other names to the Proust's law as, the law of composition or definite proportions or constant compositions.

This can understood from given example like: Oxygen is composed of 8/9 of the mass of any sample of pure water while the hydrogen fills up the remaining 1/9 of the mass. The basis of stoichiometry is structured with the law of multiple proportions along the law of definite proportions.

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If you were to be drawn into a black hole, what would happen? To the black hole, not to you.
love history [14]

Answer:

It grows

Explanation:

The blacks holes will absorb

Me hoizontally stretching me like a noodle by the spaghtification process,thus growing bigger.

7 0
3 years ago
When the displacement of a mass on a spring is 12 a the half of the amplitude, what fraction of the mechanical energy is kinetic
Sveta_85 [38]
You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).

And you know that the potential energy, PE, is [ 1/2 ] k (x^2)

Then, use x = A, to calculate the PE in the point where ME = PE.

ME = PE = [1/2] k (A)^2.

At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2

=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME

So, if PE is 1/4 of ME, KE is 3/4 of ME.

And the answer is 3/4


7 0
3 years ago
The maximum force that a grocery bag can withstand without ripping is 250 n. Suppose that the bag is filled with 20 kg of grocer
zhannawk [14.2K]

Answer:

Groceries stay in the bag.

Explanation:

Given:

Maximum force = 250 N

Bag filled with = 20 kg

Lifted acceleration = 5.0\ m/s^2

Solution:

We need to calculate the exerted force on the grocery bag by using Newton's second law.

F = ma

Where:

F = Exerted force on the object.

m = Mass of the object in kg

a = Acceleration of the object in m/s^2

Now, we substitute m = 20 kg and a = 5.0\ m/s^2 in Newton's second law,

F = 20\times 5.0

F = 100\ m/s^2

Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.

3 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:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \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
Frank has a paperclip. It has a mass of 12g and a volume of 3cm3. What is its density?​
Mkey [24]

Answer:

D = 4 g/cm³

Explanation:

Density = Mass / Volume

Step 1: Define

D = x

M = 12 g

V = 3 cm³

Step 2: Substitute

D = 12g/3 cm³

Step 3: Simplify

D = 4g / cm³

4 0
3 years ago
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