Que es la friccion <br> Cual es la primera ley de newton

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

2 years ago

Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet

kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + WW = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\$

$\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

$W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg$

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0

2 years ago

A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant

Juliette [100K]

<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, <u>as long as the spring is not permanently deformed</u>:

$F=k (x-x_{o})$ (1)

Where:

$k$ is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

$x_{o}$ is the length of the spring without applying force.

$x$ is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth $W$ of the block:

$W=m.g$ (2)

Where $m=250g=0.25kg$ is the mass of the block and $g=9.8\frac{m}{s^{2}}$ is the gravity acceleration.

$W=(0.25kg)(9.8\frac{m}{s^{2}})$ (3)

$W=2.45N$ (4)

Knowing the force applied $W$ and $x=18cm=0.18m$ and $x_{o}=0$ , we can substitute the values in equation (1) and find $k$ :

$W=k (x-x_{o})$ (5)

$2.45N=k (0.18m-0m)$ (6)

<u>Finally:</u>

$k=13.61\frac{N}{m}$

4 0

2 years ago

Light travels at a speed of close to 3 x 10^5 km/s in vacuum. Given that it takes light 8 min and 19 s to travel the distance fr

never [62]

Answer:

1497×10⁵ km

Explanation:

Speed of light in vacuum = 3×10⁵ km/s

Time taken by the light of the Sun to reach the Earth = 8 min and 19 s

Converting to seconds we get

8×60+19 = 499 seconds

Distance = Speed × Time

$\text{Distance}=3\times 10^5\times 499\\\Rightarrow \text{Distance}=1497\times 10^5\ km$

1 AU = 1497×10⁵ km

The Sun is 1497×10⁵ km from Earth

8 0

3 years ago

If the x-component of a force vector is 5.17 newtons and its y-component is 8.80 newtons, then what is its magnitude?

olya-2409 [2.1K]

Answer:

10.21 N

Explanation:

As the force is a vector, it can be decomposed in two components perpendicular each other, so there is no projection of one component in the direction of the other.

When divided in this way, the magnitude of the resultant vector can be found simply applying trigonometry, as follows:

F² = Fx² + Fy² ⇒ F = √(Fx)²+(Fy)²

Replacing by Fx= 5.17 N and Fy = 8.8 N, we get:

F = √(5.17)²+(8.8)² =10.21 N

3 0

2 years ago

Que es la friccion Cual es la primera ley de newton