The process of conversion of solid into gas and again gas into solid is known as

If a star is moving away from you at a constant speed, how do the wavelengths of the absorption lines change as the star gets fa

Minchanka [31]

Answer:

they stay shifted the same amount to the red

Explanation:

Redshift is given by

$z=\dfrac{\lambda_o-\lambda_e}{\lambda_e}$

Where,

$\lambda_o$ = Wavelength observed

$\lambda_e$ = Wavelength emitted

Also

Transverse redshift is given by

$1+z=\dfrac{1}{\sqrt{1-v^2/c^2}}$

v = Velocity of object

c = Speed of light = $3\times 10^8\ m/s$

So, if the velocity is constant the redshift remains the same

8 0

3 years ago

An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

jenyasd209 [6]

Answer:

a) i = -9.63 cm , h ’= .0.24075 cm erect

b) i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

1 / f = (n-1) (1 / R₁ - 1 / R₂)

1 / f = (1.70 -1)) 1 / ∞ - 1/13)

1 / f = 0.0538

f = - 18.57 cm

Now we can use the constructor equation

1 / f = 1 / o + 1 / i

1 / i = 1 / f - 1 / o

1 / i = -1 / 18.57 -1/20

1 / i = -0.1038 cm

I = -9.63 cm

For the height of the

image let's use magnification

m = h '/ h = - i / o

h ’= -h i / o

h ’= - 0.5 (-9.63) / 20

h ’= .0.24075 cm

b) we invert the lens

The focal length is

1 / f = (1.70 -1) (1/13 - 1 / int)

1 / f = 0.0538

f = 18.57 cm

1 / i = 1 / f -1 / o

1 / I = 1 / 18.57 - 1/20

1 / I = 3.85 10-3

i = 259.74 cm

h ’= - 0.5 259.74 / 20

h ’= 6.4935 cm

7 0

3 years ago

Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa

hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

$T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}$

Put the value into the formula

$T=1\sqrt{\dfrac{1+0.6}{1-0.6}}$

$T=2\ years$

Similarly,

The expression for distance cover by A

$D=d\sqrt{1-\dfrac{v^2}{c^2}}$

$D=12\sqrt{1-(0.6)^2}$

$D=9.6\ ly$

We need to calculate the time

Using formula of time

$t=\dfrac{D}{v}$

$t=\dfrac{9.6}{0.6}$

$t=16\ years$

We need to calculate the signals received by twin A

Using formula for number of signals

$n=\dfrac{t}{T}$

Put the value into the formula

$n=\dfrac{16}{2}$

$n=8\ signals$

Hence, 8 signals received by twin A during the trip.

7 0

3 years ago

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

So

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

8 0

3 years ago

What is a material that reduces the flow of heat by conduction, convection, and radiation?

Oksi-84 [34.3K]

The answer is a insulator

4 0

4 years ago

Read 2 more answers