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LenaWriter [7]
3 years ago
10

Two helium-filled balloons are released simultaneously at points A and B on the x axis in an earth-based reference frame. Balloo

n A is to the left of balloon B. Which one of the following statements is true for an observer moving in the +x direction?
A. The observer sees balloon B released before balloon A.
B. The observer always sees the balloons released simultaneously.
C. The observer could see either balloon released first depending on her speed and the distance between A and B.
D. The observer sees balloon A released before balloon B.
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

The correct answer is C

Explanation:

In this exercise you are asked to analyze the following situation: for a fixed observer, the two balloons are launched at the same time.

If the observer is mobile moving towards the positive side of axis ax we have several

possibilities when starting the movement

* the observer is to the left of the two balloons

* the observer is between the two balloons

* the observer is to the right of the two baloomls

The time it takes for the signal to arrive to know which ball goes first is

             v = d / t

             t = d / v

If the signal goes at the speed of light, the speed is a constant and the time will depend only on the distance, so we see that the trigger changes depending on the relative position between a given ball and the observer.

Consequently, it will be seen which comes out first, depending on the relative position with the observer.

           

The correct answer is C

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A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d
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Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

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All of the fallowing are possible sources of error in a scientific investigation exept for
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c adding research resources during an investigation

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What do scientists use to explain an atom or the universe?
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Scientists use theories to explain these things

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3 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

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3 years ago
What type of clouds are associated with low pressure?
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Cumulus and cumulonimbus<span />
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