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3 years ago

How can people take picture fpr black hole?

Physics

2 answers:

Anni [7]3 years ago

8 0

Answer:

All you have to do (if you have an camera or iphone) you just do port-rate mode

Explanation:

Jet001 [13]3 years ago

5 0

We actually cant, really. Black Holes are really very powerful and don't just happen in some random place in the cosmos. We have got shots of Black holes that you cannot even see, but all the other big black holes that look like something straight of a sci-fi movie that look real, are really just photo- shopped. good question. Hope This Helped.

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If I push a box at a constant rate is there friction force acting on it?

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Yes, the friction is acting in the opposite direction you are pushing.

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3 years ago

How do I calculate the value of the slope graph

marysya [2.9K]

Answer:

Slope is calculated using the equation: slope= rise/run (y/x). Take take y value of a point and put it over the x value of the same point and then simplify the fraction.

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3 years ago

Four main parts of a spiral galaxy

Svetlanka [38]

The 'Bulge', the 'Disk', the 'Spiral Arms', and the 'Halo' those are four parts of a spiral galaxy. Try finding out the many different types of galaxies to understand which is which.

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3 years ago

Read 2 more answers

Estimate how fast your hair grows, in units of m/s, assuming it takes 30 days for your hair to grow 1 inch. note that 1 inch =2.

VashaNatasha [74]

Answer:

1. 9.8 x 10^-7 cm/s

2. 50970 L

Explanation:

time, t = 30 days = 30 x 24 x 60 x 60 seconds = 2592000 seconds

length of hair, d = 1 inch = 2.54 cm

rate of growth = length of hair grows per second = 2.54 / 2592000

= 9.8 x 10^-7 cm/s

Thus, the grown rate of hair is 9.8 x 10^-7 cm/s.

2. length of the pool, l = 15 feet

width of the pool, w = 15 feet

height of pool, h = 8 feet

Volume of the pool, V = length x width x height

V = 15 x 15 x 8 = 1800 ft^3

To convert ft^3 into m^3 , we use

1 ft = 0.3048 m

so, 1 ft^3 = (0.3048)^3 m^3 = 0.0283 m^3

So, 1800 ft^3 = 1800 x 0.0283 = 50.97 m^3

now, 1 m^3 = 1000 L

So, 50.97 m^3 = 50.97 x 1000 L = 50970 L

3 0

3 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago