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evablogger [386]

3 years ago

13

I'm confused with this I need this ASAP

1 answer:

Gnoma [55]3 years ago

6 0

I’m pretty sure it’s a

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If forces acting on an object are unbalanced, which factor may result from an unbalanced force? The net force is negative. There

Oksi-84 [34.3K]

Answer:the Forces cancel out each other

Explanation:the forces cancel out each other

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2 years ago

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If you wanted to estimate how much money a certain appliance costs you, how would you do it?

serious [3.7K]

Interesting question.

There are a number of factors that may attribute to the pricing of an item say quality of material, number of special features available, and certainly a common notion of brand value (e.g H&M vs GUCCI). Of course, it does not necessarily mean more face value = more expensive.

Typically, to estimate, I would check the brand and material component.

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3 years ago

A speedboat travels from the dock to the first buoy, a distance of 20 meters, in 18 seconds. It began the trip at a speed of o m

MArishka [77]

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3 years ago

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An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago

Which equation describes the cosines function for right triangle?

Bogdan [553]

Answer:a

Explanation:

8 0

3 years ago