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Charra [1.4K]
3 years ago
11

How do I calculate the speed of light and the distance between the earth and the moon?

Physics
1 answer:
MAVERICK [17]3 years ago
7 0

To calculate the distance to a star, astronomers observe it from different places along Earth's orbit around the Sun. If they measure the object's position several months apart, their "two eyes" will have a separation of well over 100 million miles!

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Why do we balance chemical equations
antiseptic1488 [7]
So it could follow the correct mass for the atom
5 0
3 years ago
Give me 3 events of matter and energy that you can't explain yourself ( need some ideas)​
nalin [4]

Answer:

Plasma, Supersolid, Superfluid

Explanation:

I don't know if this is what you are looking for, but I have no idea what these even are.

8 0
3 years ago
In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom
krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2  (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R =  38.5 x ( v^2 )

77 x 9.8 x 3.65 =  38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) =  71.54

v = 8.4 m/s  ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)  

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf  

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy  = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0
4 years ago
What is the electric flux through the surface a1 that encloses all three objects?
vovangra [49]
For the answer to the question above, I assume that the question is two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3,?<span>two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3, is electrically neutral.
</span>From Gauss's law: 
<span>Flux = ∫c E . dA = q/eo </span>

<span>Since this surface encloses all </span>
<span>charge, we can simplify: </span>

<span>Flux = (q1+q2+q3)/eo </span>
<span>Flux = </span>
<span>( (1*10^-6)+(-1.9*10^-6)+(0) )/(8.85*10^-12) = -101694.92 N·m2/C</span>
6 0
3 years ago
The width of the central maximum is defined as the distance between the two minima closest to the center of the diffraction patt
Gnom [1K]

Answer:

The value of the angle is \bf{ \sin^{-1}[h/am_{e}v]}.

Explanation:

Given:

The condition for diffraction minima is

a \sin \theta = m \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

where, a is the slit-width, \theta is the angle of incidence, m is the order number and \lambda is the wavelength of the light.

The wavelength of an electron traveling through a medium is governed by de Broglie's hypothesis.

According to de Broglie's hypothesis

\lambda &=& \dfrac{h}{p}\\               &=& \dfrac{h}{m_{e}v}

Here, h is Planck's constant, m_{e} is the mass of the electron and v is the velocity of the electron.

For first minimum m = 1.

From equation (1), we have

&& a \sin \theta = \dfrac{h}{m_{e}v}\\&or,& \theta = \sin^{-1}[\dfrac{h}{am_{e}v}]

6 0
3 years ago
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