What is the kinetic energy of a 1 kg ball thrown to a height of 10 meters?

rt A For a given substance, do you expect the density of the substance in its liquid state to be closer to the density in the ga

Nat2105 [25]

Answer:

A substance in its liquid state is closer to the density of its solid phase than the density of its gaseous phase.

Explanation:

For a substance in its liquid state we can expect the density of the substance more closer to the density of its solid state than its gaseous state because the the inter-molecular space is much close near to incompressible in the liquid state and the the inter-molecular force of attraction is much higher as compared to gaseous state.

In contrast to the molecular properties in liquid state gases have almost negligible inter-molecular force of attraction and very huge inter-molecular spacing which makes it well compressible.

8 0

3 years ago

Identify each statement as an example of melting or sublimation,

xeze [42]

1st is sublimation

2and is melting

3red is melting

4th is sublimation

sublimation is just "skipping" the liquid phase / state

5 0

3 years ago

An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos

erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

$\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm$

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

$\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm$

3 0

3 years ago

What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a

viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

$COP = 8.49$

b

$COP_1 = 9.49$

Explanation:

From the question we are told that

The lower operation temperature of refrigerator is $T_1 = -8.00^oC = 265 \ K$

The upper operation temperature of the refrigerator is $T_2 = 23.2 ^oC = 296.2 \ K$

Generally the refrigerators coefficient of performance is mathematically represented as

$COP = \frac{T_1}{T_2 - T_1 }$

=> $COP = \frac{265}{296.2 - 265 }$

=> $COP = 8.49$

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

$COP_1 = \frac{T_2}{ T_2 - T_1}$

=> $COP_1 = \frac{296.2}{ 296.2 - 265 }$

=> $COP_1 = 9.49$

8 0

3 years ago

Which one of the following statements does not accurately describe vibrations?

Paladinen [302]

A. Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.

4 0

3 years ago