What force keeps our atmosphere from floating out into space?

En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

for water we have to

c = 4.18J / g ° C

replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

7 0

3 years ago

An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pi

Harman [31]

Answer:

The length is $l = 8.6 \ m$

Explanation:

From the question we are told that

The frequencies of the two successive harmonics are $f_1 = 220 \ Hz$ , $f_2 = 240 \ Hz$

The speed of sound in the air is $v_s = 343 \ m/s$

Generally the frequency of a given harmonic is mathematically represented as

$f_n = \frac{n v }{2l}$

Here n defines the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

$220 = \frac{n v}{2l}$

and

$240 = \frac{(n+1) v}{2l}$

So

$\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220$

=> $\frac{v}{2l} = 20$

=> $l = 8.6 \ m$

7 0

3 years ago

Calculate the frequency of visible light having a wavelength of 410 nm

maria [59]

Given: Wavelength λ = 410 nm convert to Meters m = 4.10 x 10⁻⁷ m

Speed of light c = 3 x 10⁸ m/s

Required: Frequency f = ?

Formula: c = λf

f = c/λ

f = 3 x 10⁸ m/s/4.10 x 10⁻⁷ m

f = 7.32 x 10¹⁴/s or 732 Thz (Terahertz)

6 0

3 years ago

100 points! please help!!!

IRINA_888 [86]

Answer:

Explanation:

A plane flies due north (90° from east) with a velocity of 100 km/h for 2 hours.

With no wind, it will be 100*2 = 200 km north of its starting point.

But a steady wind blows southeast at 30 km/h at an angle of 315° from due east.

So the wind itself will blow the plane 30*2 = 60km at an angle of 315° from due east.

That is the same as 60*cos315° = 42.43km due east and 60*sin315° = -42.43km north.

Combining, the plane is at 42.43km due east and 200-42.43 = 157.57km due north from its starting point.

7 0

3 years ago

Read 2 more answers

A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i

nexus9112 [7]

Answer:

E = k q / a² (1.3535) (- i ^ + j ^)

E = k q / a² 1.914 , θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

E₁₂ = k q / a²

On the y axis

E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

d = √ (a² + a²) = a √2

let's look for the field

E₁₃ = k q / d²

E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

cos 45 = E₁₃ₓ / E₁₃

E₁₃ₓ = E₁₃ cos 45

sin 45 = E_{13y} / E₁₃

E_{13y} = E₁₃ sin 45

E₁₃ₓ = k q / 2a² cos 45

E_{13y} = k q / 2a² sin 45

let's find each component of the electric field

X axis

Eₓ = -E₁₂ - E₁₃ₓ

Eₓ = - k q / a² - k q / 2a² cos 45

Eₓ = - k q / a² (1 + cos 45/2)

cos 45 = sin 45 = 0.707

Eₓ = - k q / a² (1 + 0.707 / 2)

Eₓ = - k q / a² (1.3535)

Y axis

E_y = E₁₄ + E_{13y}

E_y = k q / a² + k q / 2a² sin 45

E_y = k q / a² (1 + sin 45/2)

E_y = k q / a² (1.3535)

we can give the results in two ways

E = k q / a² (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

E = √ (Eₓ² + E_y²)

E = k q / a² 1.3535 √2

E = k q / a² 1.914

we use trigonometry for the angle

tan θ = E_y / Eₓ

θ = tan⁻¹ (E_y / Eₓ)

θ = tan⁻¹ (1 / -1)

θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

θ‘= 90 + 45

θ’= 135

4 0

3 years ago