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Karolina [17]
3 years ago
13

How does an atom of bromine-79 become a bromide ion with a -1 charge?

Physics
1 answer:
Tanzania [10]3 years ago
3 0

a the atom loses 1 proton to have a total of 34

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According to Newton's<br> Ist law, what will an<br> object in motion tend<br> to do?
SCORPION-xisa [38]

Answer:

According to <em>Newton's first law of motion:</em>

<u>An object in motion tends to remain in motion unless an external force acts upon it.</u>

<u>It stays in motion with the same speed and goes in the same direction.</u>

<u></u>

<em>Hope this helped </em>

<em>:)</em>

4 0
3 years ago
Read 2 more answers
A 5-turn square loop (10 cm along aside, resistance = 4.0 ) is placed in a magnetic field that makes an angle of30o with the pla
andreev551 [17]

Answer:

Explanation:

Area A of the coil = .1 x .1 = .01 m²

no of turns n = 5

magnetic field B = .5 t²

Flux  Φ perpendicular to plane passing through it.= nBA sin30

rate of change of flux

dΦ/dt = nAdBsin30 / dt

= nA d/dt (.5t²x .5 )

= nA x 2 x .25 x t

At t = 4s

dΦ/dt = nA x 2

= 5x .01 x 2

= .1

current = induced emf / resistance

= .1 / 4

= .025 A

= 25 mA.

4 0
3 years ago
Students are experimenting with circuits in their physics class and they build the two working circuits pictured below. The batt
bija089 [108]

The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs.  In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases.  In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.

<h3>What is power of the circuit?</h3>

The power of the bulb or any resistor is equal to the product of voltage and  current flowing through it.

P = VI

Circuit A has bulbs in series while the circuit B has bulbs in parallel.

When bulb 3 added to circuit A,  the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.

The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.

Thus, the last option is correct.

Learn more about power.

brainly.com/question/2933971

#SPJ1

4 0
2 years ago
Two speakers hang from racks placed in an open field. When sound of the same frequency comes from both speakers, no sound is hea
Vanyuwa [196]

Answer:

Explanation:

Wgen the sound is emitting from two speakers, the sound waves interfere each other. the locations at which the destructive interference occurs, we get no sound and the locations where constructive interference occurs, the sound occurs at that locations.

8 0
3 years ago
Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at
dybincka [34]

Answer:

The intensity at a spot 71 m away is 4.02*10^{-5}  Wm^{-2}

Explanation:

Given:

Initial intensity,I_{1}=3.0*10^{-4} Wm^{-2} at a distance, d_{1} = 26 m

Required:

New intensity, I_{2} =? at a distance, d_{2} = 71 m

Using the inverse square law,

I ∝ \frac{1}{d^{2} }

⇒I_{1}I_{1}d_{1}^{2}  =I_{2}d_{2}^{2}

I_{2} =\frac{I_{1} d_{1}^{2}  }{d_{2}^{2}} =\frac{3.0*10^{-4}*26^{2}  }{71^{2} } \\×

I_{2}=4.02*10^{-5}  Wm^{-2}

Thus, the intensity at a spot that is 71 m away is 4.02*10^{-5}  Wm^{-2}

5 0
3 years ago
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