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2 years ago

Constellation *

Physics

2 answers:

GREYUIT [131]2 years ago

7 0

Answer:

Explanation:

a group of stars forming a recognizable pattern that is traditionally named after its apparent form or identified with a mythological figure.

I hope this helps a bit.

Taya2010 [7]2 years ago

3 0

D) pattern of stars

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g A high-energy photon turns into and electron and a positron. (A positron has exactly the same mass as the electron, but opposi

yawa3891 [41]

Answer:

2 m = E / c^2 where m is mass of electron

E = h v where v is the frequency ( nu) of the incident photon

E = h c / y where y is the incident wavelength (lambda)

2 m = h / (c y)

y = h / (2 m c) wavelength required

y = 6.62 * 10E-34 / (2 * 9.1 * 10E-31 * 3 * 10E8) m

y = 3.31 / 27.3 E-11 m

y = 1.21 E -12 m = .0121 Angstrom units

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Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of

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Well they could go down a hill to gain more kinetic energy.

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2 years ago

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Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

$v'^2-u^2=2gh'$

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s$

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

$\Delta t = 2\cdot 0.04 =0.08 s$

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3 years ago

How heavy is an object that displaces 400N of water in a pool?

julia-pushkina [17]

All we can say is that the object's volume is about 41 liters. That's the same as the volume of water displaced.

We can't say anything about the object's weight. There is no direct connection between the weight of the object and the weight of the water it displaces.

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2 years ago

What is the formula to calculate the thickness of a test tube?

Maslowich

Answer:

Having the inside dimensions (ID) and the outside dimensions (OD) will allow you to figure out the wall thickness on tubing. You would need to subtract the ID from the OD and then divide by two. This number is the wall thickness.

Explanation:

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