Physics High School

A hot piece of brass (c = 380 J/kg˚C) is placed in water (c = 4180 J/kg˚C) to cool. The final temperature of the “mixture” will be closer to the initial temperature of which substance? Explain.

Answers

Answer 1

Given :

A hot piece of brass (c = 380 J/kg˚C) is placed in water (c = 4180 J/kg˚C) to cool.

To Find :

The final temperature of the “mixture” will be closer to the initial temperature of which substance.

Solution :

We know, change in temperature is inversely proportional to specific heat of the material.

In the given case specific heat of water is much greater than specific heat of brass.

So, change in temperature will be much in brass as compared to water.

Therefore, final temperature of mixture will be closer to the initial temperature of water.

Hence, this is the required solution.

Related Questions

is a directional antenna useful for mobile phones? why? how can the gain of an antenna be improved?

Answers

By focusing the signal in a specific direction, a directional antenna can increase the signal strength, providing better reception and faster data speeds. Gain of an antenna can be improved by increasing its directivity

A directional antenna can be useful for mobile phones as it can increase the signal strength and provide better reception and data speeds. The gain of an antenna can be improved by increasing its directivity, by altering its length and shape, or by adding additional elements. However, it is important to note that the improvement in gain will only be in the specific direction the antenna is facing, and signals from other directions may be weakened.

A directional antenna can be useful for mobile phones, especially in areas with weak signals. By focusing the signal in a specific direction, a directional antenna can increase the signal strength, providing better reception and faster data speeds.

For example, a Yagi antenna is a common type of directional antenna that is often used to improve the signal of a cell phone in a specific direction.

The gain of an antenna can be improved by increasing its directivity, which is the measure of how much the antenna concentrates the radio waves in a specific direction. To increase the directivity, the length and shape of the antenna can be altered, or additional elements such as reflectors or directors can be added. Increasing the size of the antenna can also increase its gain, but this may not always be practical in the case of a mobile phone.

Learn more about directional antenna here:

https://brainly.com/question/25789224

#SPJ4

A system is at rest. The mass of each object is shown. Neglecting the mass of the strings, how many forces are acting on the 6 kg block?
string 1
6kg
string 2
9kg
A. There is 1 force.
B. There are 2 forces.
C. There are 3 forces.
D. There are 4 forces.

Answers

A system is at rest. The mass of each object is shown. Neglecting the mass of the strings, the number of forces that are acting on the 6 kg block is option B. There are 2 forces

How did we arrive at the assertion?

The system is at rest, so the forces acting on the 6 kg block must be balanced. Since the only forces acting on the block are the tensions in the strings, there must be two forces acting on the block. The correct answer is B. There are 2 forces.

Therefore, the mass of each object is shown. Neglecting the mass of the strings, the number of forces that are acting on the 6 kg block is option B. There are 2 forces.

learn more about forces: https://brainly.com/question/12970081

#SPJ1

Which of the following does not describe when something has acceleration?
Changes in displacement
Changes in speed
Changes in direction
Changes in speed and direction

Answers

Answer:

changes in displacement

Explanation:

Sally has a mass of 45.9 kilograms. Earth has a mass of 5.98 x 10^24 kilograms and an average radius of 6.38 x 10^6 meters.What is the force due to gravity between Sally and Earth? Include units in your answer. Answer must be in 3 significant digits.

Answers

Newton's law of universal Gravity:

F = G * (M1 * M2)/ r^2

Where:

G = gravitational constant = 6.674 x10^-11 Nm^2kg^2

M1 = mass 1 = 45.9 kg

M2 = mass 2 = 5.98 x 10 ^24 kg

r = Distance between the 2 objects = 6.38 x 10 ^6 m

Replacing;

$F=6.674x10^{-11}Nm^2kg^2\cdot\frac{45.9\operatorname{kg}\cdot5.98x10^{24}\operatorname{kg}}{(6.38x10^6m)^2}$

F = 450.048 N

please help me out i'm so depressed and such a failure
how does the brightness of the lamp (L) vary if we dispalce the cursor (C) of the rheostat
a)Towards A
b) Towards B

Answers

Answer:

ok so first ur not a failure and its okay don't be so hard on ur self

If a pump handles a fluid at a temperature 177 C and a pressure of 164 psig with 3.66 m/s at suction nozzle. What is the NPSH (ft) available if the vapor pressure of fluid is 134 psia and specific gravity of fluid is 0.89 at 177 C?
A.80
B.90
C.120
D.105

Answers

The NPSH (ft) available is approximately 120 ft. The correct option is C.

To calculate the Net Positive Suction Head (NPSH) available, we need to use the following formula:

$NPSH (ft) = \left[\dfrac{(P - Pv)} {(\rho \times g)}\right] - \left\dfrac{V^2} { (2 g)}$

Where:

P is the absolute pressure at the suction nozzle (psia)Pv is the vapour pressure of the fluid (psia)ρ is the density of the fluid (lb/ft^3)g is the acceleration due to gravity (32.2 $\dfrac{ft}{s^2}$)V is the velocity of the fluid at the suction nozzle (ft/s)

Given:

P = 164 psig (pressure at the suction nozzle)Pv = 134 psia (vapour pressure of the fluid)ρ = specific gravity x ρ(water) (density of the fluid)= 0.89 x 62.4 $\dfrac{lb}{ft^3}$(assuming water as the reference fluid)g = 32.2 $\dfrac{ft}{s^2}$ (acceleration due to gravity)V = 3.66 $\dfrac{m}{s}$ x 3.28 $\dfrac{ft}{m}$

Calculating NPSH:

P = 164 psig + 14.7 psia (converting psig to psia)

= 178.7 psia

Calculate the density,

ρ = 0.89 x 62.4

ρ = 55.536 $\dfrac{lb}{ft^3}$

The velocity is calculated as,

V = 3.66 x 3.28

V = 12.0288 $\dfrac{ft}{s}$

$NPSH (ft) = \dfrac{(178.7 - 134 ) (55.536 * 32.2 )] - (12.0288) } { (2 \times 32.2 )}$

$NPSH (ft) = \dfrac{44.7 (1783.1712)] - (0.1888 ) }{ 64.4 }$

NPSH (ft) = 120 ft

Therefore, the NPSH (ft) available is approximately 120 ft.

To know more about NPSH follow

https://brainly.com/question/13277053

#SPJ4

A subpanel in an older house that is wired with non-metallic sheathed cable (nm) has one circuit with a grounding wire. the grounding wire:_____.

Answers

A subpanel in an older house that is wired with non-metallic sheathed cable (nm) has one circuit with a grounding wire. the grounding wire should be connected to a separate grounded buss bar.

Brass, Aluminum or Copper Bus bars are metal strips or rods. They come in a variety of shapes and sizes that determine the maximum current (ampacity) they can carry safely and reliably. This also means that it is important to install the best one for a particular application.

Its job is to receive power from the delivery point and distribute it to the output circuits. However, unlike cables, no insulation is usually required. Busbars are often arranged in separate housings. Bus bars are used as nodes or nodes in power distribution applications and can be housed in switchgear or control cabinets and bus bar enclosures.

The grounding bar is used in environments where a common grounding point or grounding point is required or required in the distribution network. One of the main advantages of brass, aluminum, or copper grounding bars is that they can be placed anywhere on the chassis or rack.

This makes installation faster and more efficient than cables. A grounded (or neutral) bus reduces the need for regular maintenance and reduces the risk of downtime. This is very important in industries where uninterrupted power distribution is essential. Telecom and power supplies are two prominent examples.

However, it is not uncommon for some neutral bus bars to have insulation.

Learn more about bus bars here : https://brainly.com/question/8496713

#SPJ4

A resistor allows a current of 0.02 A to flow through it when there is a potential difference of 10 V between its ends. What is its resistance?

Answers

Answer:

The resistance of the resistor can be calculated using Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage . Since we know that a current of 0.02 A is flowing through the resistor when there is a potential difference of 10 V , we can calculate the resistance using the following equation:

R = V/I

Where R is the resistance, V is the voltage (10 V) and I is the current (0.02 A).

Therefore, the resistance of the resistor is 500 Ω.

A canoe has a velocity of 0.38 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.60 m/s east relative to the earth find the magnitude of the velocity of the canoe relative to the river. express your answer in meters per second.

Answers

The magnitude of the velocity of the canoe relative to the river that is flowing 0.60 m/s east relative to the earth is 0.71 m/s.

The mаgnitude of the velocity of the cаnoe relаtive to the river cаn be found using the Pythаgoreаn theorem. The Pythаgoreаn theorem stаtes thаt for а right triаngle, the squаre of the hypotenuse is equаl to the sum of the squаres of the other two sides. In this cаse, the hypotenuse is the velocity of the cаnoe relаtive to the river, аnd the other two sides аre the velocity of the cаnoe relаtive to the eаrth аnd the velocity of the river relаtive to the eаrth.

Using the Pythаgoreаn theorem, we cаn find the mаgnitude of the velocity of the cаnoe relаtive to the river аs follows:

$V_{cr}$ = √($V_{ce} ^{2}$ + $V_{re} ^{2}$)

where $V_{cr}$ is the velocity of the canoe relative to the river, $V_{ce} ^{2}$ is the velocity of the canoe relative to the earth, and $V_{re} ^{2}$ is the velocity of the river relative to the earth.

Plugging in the given values, we get:

$V_{cr}$ = √((0.38 m/s)² + (0.60 m/s)²)

$V_{cr}$ = √(0.1444 m2/s² + 0.36 m2/s²)

$V_{cr}$ = √(0.5044 m²/s²)

$V_{cr}$ = 0.71 m/s

Therefore, the magnitude of the velocity of the canoe relative to the river is 0.71 m/s.

For more information about magnitude of the velocity refers to the link: https://brainly.com/question/30354282

#SPJ11

Determine the Relative Humidity, Dew Point, and Lifting Condensation Level: 1. The temperature is 30∘F and it contains 3.5 g/kg of water vapor. 2. The temperature is 50∘F and it contains 5.70 g/kg of water vapor. 3. The temperature is 70∘F and it contains 3.5 g/kg of water vapor. 4. The temperature is 80∘F and it contains 5.60 g/kg of water vapor. 5. The temperature is 80∘F and it contains 11.56 g/kg of water vapor. 6. The temperature is 30∘F and the mixing ratio is 3.5. 7. The temperature is 70∘F and the mixing ratio is 8.32. 8. The temperature is 70∘F and the mixing ratio is 3.66. 9. The temperature is 80∘F and the mixing ratio is 17.59. 10. The temperature is 50∘F and the mixing ratio is 6.54.

Answers

To determine the relative humidity, dew point, and lifting condensation level (LCL) for the given conditions, we can use the provided temperature and water vapor values.

Here are the calculations for each scenario:

1. Temperature: 30°F, Water Vapor: 3.5 g/kg - Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure)

- Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure) - LCL: N/A (Need the temperature and dew point)

2. Temperature: 50°F, Water Vapor: 5.70 g/kg

- Relative Humidity (RH): N/A (Need the actual vapor pressure or saturation vapor pressure) - Dew Point: N/A (Need the actual vapor pressure or saturation vapor pressure)