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klemol [59]
2 years ago
14

Are we real?Should we believe in hell and heaven?Should we take life seriously?Why do we feel like we are being pulled down from

our ankle quickly when we have a dream we are falling?Do we die when were asleep then come back when we wake up?Is life just a dream?Should we be scared of sleep?
Physics
2 answers:
nlexa [21]2 years ago
7 0
Uhhhhh just sleep? Are you ok?
kykrilka [37]2 years ago
6 0

Answer:

we Are real, I don’t believe hell or heaven, we should take it seriously,

Explanation:

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A slender, uniform metal rod of mass M and length lis pivoted without friction about an axis through its midpoint andperpendicul
lys-0071 [83]

Answer:

Explanation:

Moment of inertia of the metal rod pivoted in the middle

= M l² / 12

If the spring is compressed by small distance x twisting the rod by angle θ

restoring force by spring

= k x

moment of torque  about axis

= k x l /2

= k θ( l /2 )²     ( x / .5 l = θ )

=

moment of torque = moment of inertia of  rod  x angular acceleration

k θ( l /2 )²   = M l² / 12 d²θ/dt²

d²θ/dt² = 3 k/M  θ

acceleration =  ω² θ

ω² = 3 k/M

ω = √ 3 k / M

8 0
3 years ago
Read 2 more answers
If you observe an object in the universe that is roughly 8 megaparsecs in size, what is the object most likely to be?
Dmitrij [34]

E. Galaxy Cluster

Explanation:

A galaxy cluster, or cluster of galaxies, is a structure that consists of anywhere from hundreds to thousands of galaxies that are bound together by mutual gravity.

A megaparsec is a million parsecs and there are about 3.3 light years in a mega-parsec. Parsec is rather a natural distance unit for astronomers. The standard abbreviation of a mega-parsec is Mpc.

A parsec is approximately 3.09 x 1016 meters, a megaparsec is about 3.09 x 1022 meters.

Hence, 8 megaparsecs is gigantic size and that can be only of a galaxy cluster consisting of hundreds and thousands of galaxies bounded together.

Keywords: galaxies, parsec, megaparsec, galaxy cluster

Learn more about galaxy clusters and astronomical units from:

https://brainly.in/question/4624292

brainly.com/question/14214806

brainly.com/question/13315988

#learnwithBrainly

4 0
3 years ago
What is the force on a box that is being pushed to the right with 50n of force while, at the same time, being pushed to the left
elena-s [515]
I think the question meant to say net force on the box. Since force is a vector, the direction matters. 20N left is negated completely by the 50N right, which means the net force is 50N-20N to the right, 30N. 
7 0
3 years ago
A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
1. What is the potential energy of a 5.0-kg
babymother [125]

Answer:

  C.  98 J

Explanation:

The appropriate formula is ...

  PE = mgh . . . . . m is mass; below, m is meters

  PE = (5 kg)(9.8 m/s^2)(2 m) = 98 kg·m^2/s^2

  PE = 98 J

5 0
3 years ago
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