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svet-max [94.6K]
3 years ago
13

Which two of the following are the densest of the materials shown?

Physics
1 answer:
kari74 [83]3 years ago
3 0

Explanation:

The density of the material is given by,

Density = \frac{mass}{Volume}

Case 1: Rock

Density = \frac{mass}{Volume}

Density = \frac{1000}{0.5}

Density = 2000 \frac{g}{L}

Case 2: Pillow

Density = \frac{mass}{Volume}

Density = \frac{200}{10}

Density = 20 \frac{g}{L}

Case 3: Bottle of Soda

Density = \frac{mass}{Volume}

Density = \frac{2000}{2}

Density = 1000 \frac{g}{L}

Case 4: Load of bread

Density = \frac{mass}{Volume}

Density = \frac{100}{1}

Density = 100 \frac{g}{L}

Thus, rock and bottle of soda has maximum density.

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A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?
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Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where we have:

p_1 = 1.01\cdot 10^5 Pa (initial pressure is stp pressure)

V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3 is the initial volume

T_1 = 0^{\circ}=273 K is the initial temperature (stp temperature)

p_2 = 368 torr = 4.9\cdot 10^4 Pa is the final pressure

V_2 = ? is the final volume

T=155^{\circ}=428 K is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3

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3 years ago
In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a
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A. 4.64\cdot 10^{11}m

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

v=\frac{2\pi r}{T}

where we have:

v=30,000 km/s = 3\cdot 10^7 m/s is the orbital speed

r is the orbital radius

T=27 h \cdot 3600 =97,200 s is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m

B. 6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

m\frac{v^2}{r}=\frac{GMm}{r^2}

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

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M_s = 2\cdot 10^{30}kg

the mass of the black hole as a multiple of our sun's mass is

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C. 9.28\cdot 10^9 m

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Substituting numbers into the formula, we find

R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m

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