All categories

3 years ago

What type of matter is coleslaw?

1 answer:

7 0

A pure substance is a substance that contains only one type of atom or molecule. Homogeneous mixture is commonly referred to solution as there is uniform distribution all over the mixture while in a heterogeneous mixture consists of different phases or substances that are visible.
A compound is substance containing two or more elements that are chemically bound together. Therefore, coleslaw (a salad containing finely shredded cabbages and carrots and a creamy dressing) is a heterogeneous mixture.

You might be interested in

The vibrations along a transverse wave move in a direction _________.

GrogVix [38]

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

Since the oscillations is perpendicular to the direction of wave, it is a transverse wave

5 0

3 years ago

The average intensity of light emerging from a polarizing sheet is 0.775 W/m2, and that of the horizontally polarized light inci

rewona [7]

Answer:

the transmission axis of polarizing sheet makes an angle of $\Theta =26.74^{\circ}$ with the horizontal

Explanation:

We have given that intensity of light incident on the sheet $I_0=0.970W/m^2$

Average intensity of light emerging from a polarizing sheet $I=0.775W/m^2$

We have to find the angle between transmission axis with the horizontal

Intensity of light polarizing from sheet is equal to $I=I_0cos^2\Theta$

So $0.775=0.970cos^2\Theta$

$cos^2\Theta =0.798$

$cos\Theta =0.893$

$\Theta =26.74^{\circ}$

So the transmission axis of polarizing sheet makes an angle of $\Theta =26.74^{\circ}$ with the horizontal

5 0

3 years ago

A 87 kg man has a total mechanical energy of 1780 J .If he is swinging downward and is currently 1.4 m above the ground, what is

borishaifa [10]

Answer:

6.4m/s

Explanation:

The total mechanical energy of the man is 1780J.

This mechanical energy is the energy due to the motion of the body and it is a form of kinetic energy.

Also, mass = 87kg

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

1780 = $\frac{1}{2}$ x 87 x v²

v² = 40.9

v = 6.4m/s

4 0

3 years ago

Where is the word England

salantis [7]

Its in the middle near the word "FEATHERS". Ur welcome!

4 0

3 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

just olya [345]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

7 0

4 years ago