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2 years ago

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A) an electron has an initial speed of 226000 m/s. if it undergoes an acceleration of 4.0 x 1014 m/s2, how long will it take to

reach a speed of 781000 m/s? answer in units of s.

1 answer:

KIM [24]2 years ago

7 0

initial speed of 226000 m/s

acceleration of 4.0 x 1014 m/s2,

speed of 781000 m/s

What is Acceleration?

Acceleration is a rate of change of velocity with respect to time with respect to direction and speed.

A point or an object moving in a straight line is accelerated if it speeds up or slows down.

Acceleration formula can be written as,

a = (v - u ) / t m/s²

As we have to find the time taken, the formula can be altered as,

$t = \frac{v-u}{a}$

where, t - time taken to reach a final speed

v - final velocity

u - initial velocity

a - acceleration.

Substituting all the given values,

$t =\frac{781000 - 226000} {4* 1014}$

= 1.3875 × 10⁻⁹ seconds.

So, taken to reach the final speed is found to be 1.3 × 10⁻⁹ 8iH..

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3 years ago

Three children are struggling and pulling on a single toy. Two of the children, Abe and Barry, are EACH (individually) pulling w

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Solution :

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$\vec{F}_R= \vec{F}_A+\vec{F}_B$

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$|\vec{F}_R| = \sqrt{60^2+60^2}$

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$|\vec F_E| = |\vec F_R|$

$|\vec F_E| =84.853 \ N$

e). The direction of the force exerted by Eric is exactly opposite to the direction of the resultant force.

The direction of the resultant force is :

$\theta = \tan ^{-1}\left(\frac{F_y}{F_x}\right)$

$= \tan ^{-1}\left(\frac{60}{60}\right)$

= 45° north east

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3 years ago

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Aleksandr [31]

how does the electric force between two charged particles change if the distance between them is increased by a factor of 3?

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3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

lora16 [44]

Answer:

Explanation:

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4 years ago

What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about

nikitadnepr [17]

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let <em>θ</em> denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude <em>n</em>

• the car's weight with magnitude <em>w</em>

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

<em>a</em> = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ <em>F</em> (⟂) = <em>N</em> + <em>W</em> (⟂) = <em>m</em> <em>a</em> (⟂)

and

∑ <em>F</em> (//) = <em>W</em> (//) = <em>m a</em> (//)

Let the direction of <em>N</em> be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle <em>θ</em> with the banked curve, and <em>W</em> makes the same angle with the negative perpendicular axis, so that the equations above reduce to

<em>N</em> - <em>m g</em> cos(<em>θ</em>) = <em>m</em> <em>a</em> sin(<em>θ</em>)

and

<em>m g</em> sin(<em>θ</em>) = <em>m a</em> cos(<em>θ</em>)

The second equation is all we need at this point to find the ideal <em>θ</em>. The mass <em>m</em> cancels out, and we can solve for <em>θ</em> to get

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→ <em>θ</em> ≈ 3.52°

8 0

3 years ago