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3 years ago

6

if a star 100 light years from earth is beginning to expand into a giant star how long will it take for astronomers to observe t

his development explain

1 answer:

VikaD [51]3 years ago

7 0

Answer:

100years later

Explanation:

Because the lights will arrive at world after 100 years later.

You might be interested in

Eardrum vibrations are transmitted to the cochlea by a piston consisting of

Lena [83]

Answer:

ossicles (comprising of hammer, anvil, and stirrup)

Explanation:

The hammer, anvil and stirrup are also known as the malleus, incus, and stapes, respectively which are the part of the middle ear.

They transmit the sound vibrations from eardrum to cochlea by a level mechanism. The generate mechanical advantage by multiplying the force over a shorter distance from the input of a smaller force at a larger distance.

4 0

3 years ago

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

3 years ago

During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electr

WINSTONCH [101]

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

$E = \frac{\Delta V}{d}$

<u>Where:</u>

E: is electric field

ΔV: is the potential difference = 3.95 mV

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

$E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m$

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

7 0

2 years ago

A person doing chin-up weighs 700.0 N, disregarding the weight of the arms. During the first 25.0 cm of the lift, each arm exert

dlinn [17]

First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton

Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons

Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction)

Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg

Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2

Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters

Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec

Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec

4 0

3 years ago

Read 2 more answers

Why is the overall charge of the atom neutral or zero?

MrRissso [65]

Answer:

B

Explanation:

this is because the neutrons do not have a charge, the things that have charge in an atom are electrons and protons.

and in the nucleus of an atom, there are protons and neutrons so you can see that A is not the answer

if you see the periodic table, you will know that the number of electrons and protons are equal, so the charges cancel each other out, hence the charge of an atom will be neutral

let me give you a tip which I got from my teacher, never write there is no charge in the atom, this suggests that there is no protons or electrons.

instead, write, the it is neutral

hope it helps if not please report it so that someone else gets to try it out

7 0

3 years ago