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3 years ago

9

A cone with a mass of 6 g has a radius measuring 5 cm and a height of 2 cm, making the volume 52.33g/cm3 (3.14*52(2/3)). What is

its density?
A- 60 g/cm3
B- 52.33 g/cm3
C- 0.11 g/cm3

1 answer:

marysya [2.9K]3 years ago

5 0

The volume of a cone is equal to (pi*r^2*h)/3

(pi*25*2)/3= 52.35987...

Density = mass/volume

d= 6 grams/ 52.35987 cm^3= 0.11459...

Final answer: D

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Which description does the phrase "mechanical advantage" most closely relate to?

yKpoI14uk [10]

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3 years ago

Gold has a density of 19.32 g/cm 3 . If you cut a piece of gold in half, the density of each piece will be

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Answer:

See Explanation

Explanation:

The density of a substance is an intrinsic property. Intrinsic properties are properties that are characteristic of a material. It does not depend on the amount of substance present.

Since density is an intrinsic or intensive property, If you cut a piece of gold in half, the density of each piece will still be 19.32 g/cm 3 because density does not depend on the size of the gold piece.

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3 years ago

7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr

Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

$d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m$

$d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m$

From the continuity equations in pipes we have to

$A_1V_1 = A_2 V_2$

Where,

$A_{1,2}$ = Cross sectional Area at each section

$V_{1,2}$ = Flow Velocity at each section

Then replacing we have,

$(\pi r_1^2) v_1 = (\pi r_2^2) v_2$

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$v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1$

From Bernoulli equation we have that the change in the pressure is

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$7300= 8919.01 v_1^2$

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6 0

4 years ago

(a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasona

antoniya [11.8K]

Answer:

Part a)

$V = 7.2 \times 10^{11} Volts$

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

Explanation:

Part a)

As we know that the potential near the surface of metal sphere is given by the equation

$V = \frac{kQ}{R}$

here we have

Q = 8 C

R = 10.0 cm

now we have

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Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

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3 years ago

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3 years ago