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3 years ago

When you observe a distant galaxy whose photons have traveled for 10 billion years before reaching earth we are seeing that gal?

1 answer:

8 0

In reality we don't see the galaxy we see it's reflection .. the light hits or got emitted by the star travel all the way long to hit our eyes .. we see their reflection . everything around you that you see is it's reflection

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The greatest ocean depths on the earth are found in the marianas trench near the philippines. calculate the pressure due to the

tensa zangetsu [6.8K]

First, let us derive our working equation. We all know that pressure is the force exerted on an area of space. In equation, that would be: P = F/A. From Newton's Law of Second Motion, force is equal to the product of mass and gravity: F = mg. So, we can substitute F to the first equation so that it becomes, P = mg/A. Now, pressure can also be determined as the force exerted by a fluid on an area. This fluid can be measure in terms of volume. Relating volume and mass, we use the parameter of density: ρ = m/V. Simplifying further in terms of height, Volume is the product of the cross-sectional area and the height. So, V = A*h. The working equation will then be derived to be:

P = ρgh

This type of pressure is called the hydrostatic pressure, the pressure exerted by the fluid over a known height. Next, we find the literature data of the density of seawater. From studies, seawater has a density ranging from 1,020 to 1,030 kg/m³. Let's just use 1,020 kg/m³. Substituting the values and making sure that the units are consistent:

P = (1,020 kg/m³)(9.81 m/s²)(11 km)*(1,000 m/1km)
P = 110,068,200 Pa or 110.07 MPa

6 0

3 years ago

At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6

Elis [28]

The particle has constant acceleration according to

$\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k$

Its velocity at time $t$ is

$\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du$

$\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t$

$\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k$

Then the particle has position at time $t$ according to

$\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du$

$\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k$

At at the point (3, 6, 9), i.e. when $t=0$ , it has speed 8, so that

$\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64$

We know that at some time $t=T$ , the particle is at the point (5, 2, 7), which tells us

$\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}$

and in particular we see that

$v_{0y}=-2v_{0x}$

and

$v_{0z}=-v_{0x}$

Then

${v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3$

$\implies v_{0y}=\mp\dfrac{8\sqrt6}3$

$\implies v_{0z}=\mp\dfrac{4\sqrt6}3$

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

$\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k$

4 0

3 years ago

4. Ano ang matinding suliranin ang kakaharapin kapag patuloy ang pagkasira ng

jasenka [17]

Answer:

I think its D

Explanation:

Kasi Pinoy ako

5 0

3 years ago

Read 2 more answers

a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream cur

Doss [256]

Refer to the diagram shown below.

Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.

The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.

The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s

The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°

Answer: 10.2 m/s at 21.5° downstream.

7 0

3 years ago

Read 2 more answers

ARRANGE THE STEPS IN ORDER TO DESCRIBE WHAT HAPPENS TO A GAS WHEN IT COOLS

vredina [299]

What happens to has when it cools is ...
Step 1) They will start to form liquids, like condensation.
Step 2) As liquids cool, they will turn into solids.
Step 3) As solids cool, they become more stable and solid.

3 0

3 years ago