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Mademuasel [1]
3 years ago
5

The "zone of avoidance" a) Accounts for the absence of any planet between the orbits of Mars and Jupiter Prevents collisions bet

ween planets in the Solar System. b) bscures visual observation through the plane of the Milky Way d) Is the minímum distance an object must have from a black hole to avoid gravitational capture The regions that remained barren during evolution of life on Earth. e)
Physics
1 answer:
shusha [124]3 years ago
7 0

Answer:

b) obscures visual observation through the plane of the Milky Way.

Explanation:

The "zone of avoidance"

Avoidance zone, an area characterized by an apparent lack of galaxies near the Milky Way Galaxy plane and caused by the mysterious impact of interstellar dust. The zone of avoidance is completely a local Milky Way Galaxy effect.

The correct answer is

b) obscures visual observation through the plane of the Milky Way.

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irinina [24]

Answer:

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Explanation:

5 0
3 years ago
What is meant by a transverse wave?
lianna [129]
In a transverse wave the particle displacement is perpendicular to the direction of wave propagation. The animation below shows a one-dimensional transverse plane wave propagating from left to right. The particles do not move along with the wave; they simply oscillate up and down about their individual equilibrium positions as the wave passes by. Pick a single particle and watch its motion.

The S waves (Secondary waves) in an earthquake are examples of Transverse waves. S waves propagate with a velocity slower than P waves, arriving several seconds later.

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4 0
3 years ago
A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the
Elina [12.6K]

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

          T = 2\pi  \sqrt{ \frac{L}{g} }

          T² =  4\pi ^2 \frac{L}{g}4pi2 L / g

          g = 4\pi ^2   \frac{L}{T^2}

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

           g = 4\pi ^2  \frac{0.823}{1.823^2}4 pi2 0.823 / 1.823 2

            g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

             T = t / n

             ΔT = \frac{dT}{dt} Δt + \frac{dT}{dn} ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

              ΔT = Δt / n

              ΔT = Δt

now let's look for the uncertainty of g

             Δg = \frac{dg}{dL} ΔL + \frac{dg}{dT}  ΔT

             Δg = 4\pi ^2 \frac{1}{T2}   ΔL + 4π²L  (-2  T⁻³) ΔT

           

a more manageable way is with the relative error

             \frac{\Delta g}{g}   = \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}

we substitute

              Δg = g ( \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation  

let's calculate

               Δg = 9.7766 (\frac{0.001}{0.823} + \frac{1}{2}  \ \frac{0.671}{1.823})

               Δg = 9.7766 (0.001215 + 0.0184)

               Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

                Δg = 0.2 m / s2

therefore the correct result is

               g ±Δg = (9.8 ± 0.2) m / s²

5 0
3 years ago
I WILL MARK AS BRAINLIST
puteri [66]

Explanation:

Area=1.5(1.5)=2.25m^2

Force of gravity=10N

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}

⟼Pressure=

Area

Force

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}

⟼Pressure=

2.25

10

\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}

⟼Pressure=4.4Pa

5 0
3 years ago
A car with a velocity of 22 m/s is accelerated at a rate of 1.6m/s2 for 6.8s. determine the final velocity
Ivahew [28]

A car with a velocity of 22 m/s is accelerated at a rate of 1.6 m/s^2 for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

time, t = 6.8 s

acceleration, a = 1.6 m/s^2

We will be using the equation of motion:

v = u + at

\therefore v=22+1.(6.8)

\Rightarrow v=22+10.88

\Rightarrow v=32.88 \ m/s

The final velocity become 32.88 m/s.

To learn more about Attention here:

https://brainly.in/question/10557838

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3 0
2 years ago
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