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3 years ago

Summarize the Venn diagram.

Physics

1 answer:

Elena L [17]3 years ago

5 0

Answer:

If this is a multiple choice question, then the answer is D.

Explanation:

Compounds are composed of atoms, which are composed of subatomic particles and consist of matter. Since they are composed of atoms rather than vice-versa, compounds cannot be found inside atoms and are not the most basic form of matter (ruling out A and C). A pure substance contains atoms, which are each composed of subatomic particles. Therefore, a pure substance must have atoms if it contains subatomic particles (ruling out B). The only answer left is D.

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What did scientists predict the big bang should have left ??

ELEN [110]

<span>radiation, hydrogen, and helium </span>

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3 years ago

Read 2 more answers

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

4 years ago

A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c

Alekssandra [29.7K]

Answer:

865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

Height (h) of the cliff = 62 m

Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

Height (h) of the cliff = 62 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

62 = ½ × 9.8 × t²

62 = 4.9 × t²

Divide both side by 4.9

t² = 62/4.9

Take the square root of both side.

t = √(62/4.9)

t = 3.56 s

Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

Horizontal distance (s) =?

s = ut

s = 243 × 3.56 s

s = 865.08 m

Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.

6 0

3 years ago

A penny has a mass of 2.50 g, a diameter of 19.55 mm, and a thickness of 1.55 mm. Calculate the density of the material from whi

Dmitry [639]

Density = (mass) divided by (volume)

We know the mass (2.5 g). We need to find the volume.

The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.

Volume = (π) (9.775 mm)² (1.55 mm)

   = (π) (95.55 mm²) (1.55 mm)

   = (π) (148.1 mm³)

   = 465.3 mm³

We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:

   mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .

Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.

   (0.0054 g/mm³) x (10 mm / cm)³

   = (0.0054 x 1,000) g/cm³

   = 5.37 g/cm³ .

This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.

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3 years ago

Which classification scheme is based on the brightness of a star as it appears to you in the night sky?

anastassius [24]

The apparent magnitude scale is a classification scheme which is based on the brightness of stars. The range of brightness values is from 1 to 6.
The stars which are the most brightest are ranked as number 1 and also called first magnitude stars, stars which are little dimmer than number 1 are ranked as number 2 and also called second magnitude stars. Similarly the most faintest stars are ranked number 6 and also called as the sixth magnitude stars.

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3 years ago