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Dvinal [7]
2 years ago
14

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find w

ays to live much, much longer?
Physics
2 answers:
kolezko [41]2 years ago
8 0

Answer:

Given the information that PSR B1620-26 b is the oldest planet in the Milky Way, it is estimated to be about 13 billion years old, the planet is approximately 12,390 light-years away from Earth, this means that it would take 460,908,000 years to arrive at the ancient planet. If we want to reach this planet, we are going to need to make some gig “wrinkle in time” discoveries, or find ways to live much longer, considering that the current lifespan of humans is only 79 years.

Ira Lisetskai [31]2 years ago
5 0

Answer:

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

Explanation:

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

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(URGENT 75 points)You are given two same party balloons (one filled with helium and the
skad [1K]

\frac{100}{0.214+-0.006} = 454.55 g/cm3

I'm not too sure since the graduated cylinder was missing and I really don't know how to do it then. But give this a shot. Are you sure it wasn't a graduated cylinder, because I have no idea what that means

5 0
3 years ago
How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?
astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of  0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J

So, the magnitude of work done in lifting the book is 6 joules.

7 0
3 years ago
A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maxi
kozerog [31]

Answer:

6.51 m and 37.13 m

Explanation:

from the question we were given

mass of skeet = 0.25 kg

speed of skeet = 25 m/s

angle = 30 degrees to the horizon

mass of pellet = 15 g

speed of pellet = 2000 m/s

without being hit by the pellet, the x and y components of the skeet velocity are  

Vx = 25 cos 30 = 21.65

Vy = 25 sin 30 = 12.5

now

V = U + (a x t)

where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance

-25 sin 30 = 25 sin 30 + (-9.8 x t)

-12.5 = 12.5 - 9.8 t

t = 2.55 s

also

V^2 = U^2 + 2as  ( s = vertical distance and V = 0 )

0 = (25 sin 30)^2 + 2 x (-9.8) x Y

19.6 Y = 156.25

Y =7.97 m

the distance traveled without the pellet hitting the skeet can be gotten using distance = speed x time

distance = 21.65 x 2.55 = 55.2 m

applying the conservation of linear momentum

on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx)  ...equation 1

on the y axis :   (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

 Vx = 20.42m/s

0 + (0.015 x 200) = (0.25 + 0.015) Vy

 Vy = 11.32 m/s

now V^2 = U^2 + 2 as

 0 = 11.3^2 + (2 x (-9.8) x s)

s = 6.51 m                          

  • to find the extra distance moved after collision we apply

s = ut + 1/2 at^2

-7.98 = 11.32t + 1/2 (-9.8) t^2

4.9 t^2 - 11.32t + 7.98

t =  3.17 s

recall that distance = speed x time

distance = 20.42 x 3.17 = 64.73 m

the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2

= 27.6 m

therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

5 0
3 years ago
If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star
torisob [31]
There's no such thing as "stationary in space".  But if the distance
between the Earth and some stars is not changing, then (A) w<span>avelengths
measured here would match the actual wavelengths emitted from these
stars. </span><span>

</span><span>If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.

</span>In order to decide what's actually happening, and how that star is moving, 
the trick is:  How do we know the actual wavelengths the star emitted ?


 
7 0
3 years ago
If one of two interacting charges is doubled, the force between the charges will _____________.
malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

F=\frac{k q1 q2}{r^{2}}

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get F'=\frac{k (2q1) q2}{r^{2}}

F'= 2 F

Thus the force gets doubled.

4 0
3 years ago
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