Ask question

Ask question

All categories

lorasvet [3.4K]

2 years ago

14

when heated the temperature of a water sample increased from 15°C to 39°C. Is absorbed 41840 joules of heat. what is the mass of

the sample

1 answer:

Hunter-Best [27]2 years ago

6 0

Answer:

Mass of water: 43 g

Explanation

hope it helps

You might be interested in

The frequency of a microwave is 1.2 x 10^9 hertz. what is the wavelength of the given problem.

Olenka [21]

Answer:

0.25 m

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field, oscillating in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

$c=3.0\cdot 10^8 m/s$

Microwave is an example of electromagnetic waves.

The relationship between wavelength and frequency for an electromagnetic wave is:

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

f is the frequency

For the microwave in this problem,

$f=1.2\cdot 10^9 Hz$

So its wavelength is

$\lambda=\frac{3.0\cdot 10^8}{1.2\cdot 10^9}=0.25 m$

7 0

4 years ago

Which of these ideas was part of the earliest model of the atom?

babymother [125]

I think the answer is D

4 0

3 years ago

Read 2 more answers

Streak is a reliable identifier of a mineral. true or false

inysia [295]

Answer:

true?

Explanation:

Im positive but not 100% sure wait for someone else to answer and see if they say the same.

8 0

4 years ago

Read 2 more answers

A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)

Where $\epsilon$ is the body's emissivity

(the value we want to find)

Isolating $\epsilon$ from (2):

$\epsilon=\frac{P}{\sigma A T^{4}}$ (3)

Solving:

$\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}}$ (4)

Finally:

$\epsilon=0.17$ (5) This is the body's emissivity

3 0

3 years ago

A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms va

natima [27]

Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor $C= 2.00\mu F$

Current = 0.680 A

We need to calculate the inductance of the inductor

Using formula of current

$I = \dfrac{V}{Z}$

$Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}$

Put the value of Z into the formula

$I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}$

Put the value into the formula

$0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}$

$L=35.8\ mH$

Hence, The inductance of the inductor is 35.8 mH

4 0

3 years ago

Read 2 more answers