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3 years ago

1. Which of the following best describes a nebula?

Physics

1 answer:

Gekata [30.6K]3 years ago

8 0

Answer:

Explanation:

denser clouds of particles found in the interstellar medium

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Please help on this one?

telo118 [61]

i believe the answer is C.

5 0

3 years ago

What is the mechanical advantage of this system? А 2 B 3 C 4

faltersainse [42]

I think the answer will be B tell me if it’s right after

4 0

2 years ago

A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

Lerok [7]

The density of the substance is 10.5 g/cm³.

The jewelry is made out of Silver.

Density ρ is defined as the ratio of mass m of the substance to its volume V. The cylinder contains a volume V₁ of water and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.

The volume V of the jewelry is given by,

$V=V_2 -V_1$

Substitute 48.6 ml for V₁ and 61.2 ml for V₂.

$V=V_2 -V_1\\ =61.2 ml -48.6 ml\\ =12.6 ml$

calculate the density ρ of the jewelry using the expression,

$\rho =\frac{m}{V}$

Substitute 132.6 g for m and 12.6 ml for V.

$\rho =\frac{m}{V}\\ =\frac{132.6 g}{12.6 ml} \\ =10.5 g/ml$

Since $1 ml=1 cm^3$ ,

The density of the jewelry is 10.5 g/cm³.

From standard tables, it can be seen that the substance used to make the jewelry is silver, which has a density 10.5 g/cm³.

4 0

3 years ago

An ideal gas is enclosed in a piston, and 1600 J of work is done on the gas. As this happens, the internal energy of the gas inc

Phantasy [73]

Answer:

- 1100 J heat flows out

Explanation:

dW = - 1600 J (as work is done on the gas)

dU = 500 J

dQ = ?

According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

3 0

3 years ago

Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t

Step2247 [10]

Answer:

$\Delta p = 90.7 kPa$

Explanation:

specific gravity of oil is $= \frac{\rho_{oil}}{\rho_w}$

$\rho_{oil} = 0.85*1000 = 850 kg/m3$

we know that

change in pressure for oil is given as

$\Delta p = \rho gh$

here density and h is for oil

$\Delta p = 850*5 *9.81 = 41,692.5 kPa$

change in pressure for WATER is given as

$\Delta p = \rho gh$

here density is for water and h is for water

$\Delta p = 1000*5 *9.81 = 49,050 kPa$

pressure change due to both is given as

$\Delta p = 41692.3 + 49050 = 90742.5 N/m2$

$\Delta p = 90.7 kPa$

8 0

3 years ago

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