Please help.....................................

distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm.

a_sh-v [17]

Answer:

The speed of the galaxy relative to the Earth is $3.09\times 10^6\ m/s$ .

Explanation:

We have,

(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.

(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :

$v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s$

So, the speed of the galaxy relative to the Earth is $3.09\times 10^6\ m/s$ .

4 0

3 years ago

Energy that is passed on from one trophic level to the next is called what?

yan [13]

Energy decreases as it moves uptrophic levels because energy is lost as metabolic heat when the organisms from one trophic level are consumed by organisms from the next level.Trophic level transfer efficiency (TLTE) measures the amount of energy that is transferred between trophic levels.

8 0

3 years ago

GIVING BRAINLY AND THIS IS A BASKETBALL QUESTION

luda_lava [24]

Answer:

Jump Ball; used to start the game. The referee throws the ball up in the air, above two members of opposite teams, who have to jump up to try and tap the ball towards one of their teammates.

Assist: a pass made from one player to another, that directly leads to making a basket. For a pass to count as an assist, the ball has to be passed directly to the played who scores, and points must be earned.

Steal: when a defensive player intercepts the ball from an offensive player.

Lay Up: a shot made from below, by running up to the basket from the side, jumping, and bouncing the ball off of the backboard, into the basket.

Technical Foul: a foul given for "unsportsmanlike behaviour", such as making fun of an opponent.

5 0

3 years ago

Read 2 more answers

A 6 gram coin is which is initially at rest is dropped from the observation deck of a skyscraper 300 meters above the street bel

steposvetlana [31]

Answer:

a) The work done by gravity on the coin as it falls is 17.653 joules.

b) The kinetic energy of the coin at a speed of 60 meters per second is 10.8 joules.

c) The work lost due to air resistance is 6.853 joules.

d) The average force exerted on the coin due to air resistance as it fell is 0.023 newtons.

Explanation:

a) We must remind that situation with Earth-coin system must be represented by Principle of Energy Conservation and Work Energy Theorem. According to this latter, work done by gravity equals to the change in gravitational potential energy:

$\Delta U_{g} = m\cdot g \cdot \Delta z$ (Eq. 1)

Where:

$\Delta U_{g}$ - Change in gravitational potential energy, measured in joules.

$m$ - Mass, measured in kilograms.

$g$ - Gravitational accelerations, measured in meters per square second.

$\Delta z$ - Height of the skyscraper, measured in meters.

If we know that $m = 0.006\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta z = 300\,m$ , the work done gravity on the coin is:

$\Delta U_{g} = (0.006\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (300\,m)$

$\Delta U_{g} = 17.653\,J$

The work done by gravity on the coin as it falls is 17.653 joules.

b) By definition of translation kinetic energy, we get the following model:

$K = \frac{1}{2}\cdot m\cdot v^{2}$

Where:

$K$ - Kinetic energy of the coin right before hitting the street, measured in joules.

$v$ - Speed of the coin, measured in meters per second.

If we get that $m = 0.006\,kg$ and $v = 60\,\frac{m}{s}$ , the kinetic energy at this speed is:

$K = \frac{1}{2}\cdot (0.006\,kg)\cdot \left(60\,\frac{m}{s} \right)^{2}$

$K = 10.8\,J$

The kinetic energy of the coin at a speed of 60 meters per second is 10.8 joules.

c) The work lost due to air resistance is obtained derived from Principle of Energy Conservation and Work-Energy Theorem:

$W_{lost} = \Delta U_{g}-K$ (Eq. 2)

Where $W_{lost}$ is the work lost due to air resistance, measured in joules.

If we know that $\Delta U_{g} = 17.653\,J$ and $K = 10.8\,J$ , the work lost due to air resistance is:

$W_{lost} = 17.653\,J-10.8\,J$

$W_{lost} = 6.853\,J$

The work lost due to air resistance is 6.853 joules.

d) The average force exerted on the coin due to air resistance can be determined by applying definition of work, as air resistance force was antiparallel to the displacement of the coin. That is:

$W_{lost} = F\cdot \Delta z$

$F = \frac{W_{lost}}{\Delta z}$ (Eq. 3)

Where $F$ is the average force exerted on the coin due to air resistance, measured in newtons.

If we know that $W_{lost} = 6.853\,J$ and $\Delta z = 300\,m$ , then the average force exerted on the coin is:

$F = \frac{6.853\,J}{300\,m}$

$F = 0.023\,N$

The average force exerted on the coin due to air resistance as it fell is 0.023 newtons.

5 0

3 years ago

Which condition must exist in order for conduction to occur between two substances

motikmotik

Heat!!!!!!!!!!!!!!!!!!! sorry about the exclamation marks but it wont let me post in less it has more than 20 characteristics

6 0

3 years ago