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3 years ago

In a isovolumetric process the is constant.

Physics

1 answer:

V125BC [204]3 years ago

3 0

Answer:

An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0.

Explanation:

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La estrella mas proxima a la tierra esta a 2 años de luz. Calcula esta distancia en unidades del SI

tino4ka555 [31]

Answer:

$2\ ly=1.89\times 10^{16}\ m$

Explanation:

The question says that, "The closest star to the earth is 2 light years away. Calculate this distance in SI units".

Given that,

The distance between the closest star and the Earth is 2 light years.

The SI unit of distance is m. It means we need to convert light years to meters. We know that,

$1\ ly=9.461\times 10^{15}\ m$

2 light-years means,

$2\ ly=2\times 9.461\times 10^{15}\ m\\\\=1.89\times 10^{16}\ m$

So, the required distance is equal to $1.89\times 10^{16}\ m$ .

5 0

3 years ago

If, in

STatiana [176]

Answer:

a) λ = 121.5 nm , b) 102.6, 97, 91.1 nm

Explanation:

Bohr's model describes the energy of the hydrogen atom

$E_{n}$ = k² e² / 2m (1 / n²)

A transition occurs when the electron passes from n level to a lower one

$E_{i}$ - $E_{n}$ = k² e² / 2m (1 / $n_{i}$ ² - 1 / $n_{f}$ ²)

Planck's relationship is

E = h f = h c / lam

hc /λ = k² e²/ 2m(1 / $n_{i}$ ² - 1 / $n_{f}$ ²)

1 / λ = [k² e² / 2m h c] (1 / $n_{i}$ ² - 1 / $n_{f}$ ²)

1 /λ = Ry] (1 / $n_{i}$ ² - 1 / $n_{f}$ ²)

a) the first element of the series occurs for $n_{f}$ = 2

1 / λ = 1.097 10⁷ (1- 1/2²)

1 / λ = 1.097 10⁷ (1- 0.25)

1 / λ = 0.82275 10⁷

λ = 1.215 10⁻⁷ m

λ = 1,215 10⁻⁷ m (10⁹nm / m)

λ = 121.5 nm

b) the next elements of the series occur to

$n_{f}$ $n_{i}$ 1 /λ λ (10-7m) λ (nm)

3 1 1,097 10⁷ (1-1 / 9) 1,0255 102.6

4 1 1,097 10⁷ (1-1 / 16) 0.9723 97.2

∞ 1 1,097 10⁷ (1 - 0) 0.91158 91.1

5 0

3 years ago

we measure a voltage difference of 5.0 V between two points on the conducting paper between two parallel conducting electrodes.

Alex17521 [72]

Answer:

E=1824.81 V/m

Explanation:

Given that

Voltage difference = 5 V

Distance ,D= 3 mm

θ = 24°

As we know that electric filed given as

$E=\dfrac{V}{d}$

Given that D is 24° with respect to the perpendicular to the electrodes.So we have to take cos component of D.

d= D cosθ

d= 3 cos24°

d = 2.74 mm

$E=\dfrac{V}{d}$

$E=\dfrac{5}{2.74\times 10^{-3}}$

E=1824.81 V/m

8 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

What is the second cooler layer of ocean water?

maw [93]

The surface zone :)))))))))

8 0

3 years ago