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Ad libitum [116K]

2 years ago

6

Write a short paragraph that explains the central idea of the article. Use at least two details from the article to support your

response. Write a short paragraph that explains the central idea of the article. Use at least two details from the article to support your response.

1 answer:

zhannawk [14.2K]2 years ago

3 0

Answer:

you should write about a book you read

Explanation:

because maybe you got really good things in it

or here is an example

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What is the value of sin r and sin i in physics

taurus [48]

Answer:

★The second law of refraction

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a light of given colour and for a given pair of media. This law is also called Snell's law of refraction. If 'i' is the angle of incidence and 'r' is the angle of refraction then, Sin i/Sin r = constant

This constant value is called the refractive index of the second medium with respect to the first.

7 0

3 years ago

A system is best defined as?

erma4kov [3.2K]

Answer:

a group of organs working together to perform a function.

Explanation:

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3 years ago

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To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

iren [92.7K]

Answer:

$372.3 J/^{\circ}C$

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

$P=VI=(3.6)(2.6)=9.36 W$

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

$E=Pt=(9.36)(350)=3276 J$

Finally, the change in temperature of an object is related to the energy supplied by

$E=C\Delta T$

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

$\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C$ is the change in temperature

Solving for C, we find:

$C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C$

5 0

2 years ago

You weigh 716 newtons on Earth. You

Otrada [13]

Answer:

537 N

Explanation:

The force due to gravity of a planet is:

F = GMm / r²

where G is the universal gravitational constant

M is the mass of the planet

m is the mass of the object

and r is the distance between the object and the center of the planet

On Earth, you weigh 716 N, so:

716 N = GMm / r²

On planet X:

F = G (3M) m / (2r)²

F = 3/4 GMm / r²

F = 3/4 (716 N)

F = 537 N

8 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago