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2 years ago

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A foam ball of mass 0.150 g carries a charge of -2.00 nC. The ball is placed inside a uniform electric field, and is suspended a

gainst the force of gravity. What are the magnitude and direction of the electric field?

1 answer:

brilliants [131]2 years ago

3 0

Answer:

Electric field, $E=-7.35\times 10^5\ N/C$ in the direction of gravity.

Explanation:

Given that,

Mass of the ball, m = 0.15 g

Charge on the ball, $q=-2\ nC=-2\times 10^{-9}\ C$

The ball is placed inside a uniform electric field, and is suspended against the force of gravity such that,

$mg=qE$

$E=\dfrac{mg}{q}$

$E=\dfrac{0.15\times 10^{-3}\times 9.8}{-2\times 10^{-9}}$

$E=-7.35\times 10^5\ N/C$

The electric field will be acting in the direction of gravity. Hence, this is the required solution.

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The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em> $1.83\times 10^{7}\,kg$ <em>. The wheels of the locomotive push back on the tracks with a constant net force of </em> $7.50\times 10^{5}\,N$ <em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

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2 years ago

The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me

KIM [24]

As per given equation we have

$x = bt^u + ct^v$

now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation

now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

$bt^u = M^0L^1T^0$

$b*T^8 = M^0L^1T^0$

$b = M^0L^1T^{-8}$

similarly for other term we have

$ct^v = M^0L^1T^0$

$c*T^7 = M^0L^1T^0$

$c = M^0L^1T^{-7}$

<em>so above are the dimensions of b and c</em>

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2 years ago

A car is traveling in a race.The car went from initial velocity of 35m/s to the final velocity of 65m/s in 5 seconds what was th

I am Lyosha [343]

Acceleration is the change in velocity divided by time. The change in velocity is -30m/s and time is 5s. If you divide -30m/s by 5s, you get -6m/s<span>².</span>

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Carl and Abby are discussing what they have learned in class about electricity and magnetism. Carl tells Abby that an electric c

Vikki [24]

Both? I'm not too sure

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2 years ago

If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over

astra-53 [7]

Let me give you the procedure like this:
Lets say that F is the fraction of the rope hanging over the table
If its like that then we have to take into account that the <span>friction force keeping on table is given by the following formula:</span>
<span>Ff = u*(1-f)*m*g </span>
and we need to know aso that <span>gravity force pulling off the table Fg is given by this other formula:</span>
<span>Fg = f*m*g </span>
What you need to do is <span>Equate the two and solve for f: </span>

<span>f*m*g = u*(1-f)*m*g </span>
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2 years ago

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