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3 years ago

6

Заряди +10 нКл і +18нКл розташовані на відстані 12 см один від одного.Яка сила діятиме на заряд +2 нКл,який розміщений на прямій

,що з'єднує заряди на відстані 3 см від першого заряду С малюнком

1 answer:

ludmilkaskok [199]3 years ago

3 0

Hdjdjdjdhfhdbdhdbdbdhdhdhdhdhdhdgdgdhd

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3 years ago

Read 2 more answers

Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut

Harman [31]

Answer: a) 11.76 m/s b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

$V_{f}=V_{o}+at$ (1)

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (2)

Where:

$V_{f}$ Is the final velocity of the object

$V_{o}=0$ Is the initial velocity of the object (it was dropped)

$a=9.8 m/s^{2}$ is the acceleration due gravity

$d$ is the height of the tower

$t=0.02min=1.2 s$ is the time it takes to the object to reach the ground

b) Begining with (1):

$V_{f}=0+at$ (3)

$V_{f}=at=(9.8 m/s^{2})(1.2 s)$ (4)

$V_{f}=11.76 m/s$ (5) This is the final velocity of the object

a) Substituting (5) in (2):

$(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d$ (6)

Clearing $d$ :

$d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})}$ (7)

$d=7.056 m$ (8) This is the height of the tower

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3 years ago

The position x, in meters, of an object is given by the equation:

LenaWriter [7]

Answer:

The SI units of A, B and C are :

$m,\ m/s\ and\ m/s^2$

Explanation:

The position x, in meters, of an object is given by the equation:

$x=A+Bt+Ct^2$

Where

t is time in seconds

We know that the unit of x is meters, such that the units of A, Bt and $Ct^2$ must be meters. So,

$A=m$

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$b=\dfrac{m}{s}=m/s$

$Ct^2=m$

$C=m/s^2$

So, the SI units of A, B and C are :

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So, the correct option is (B).

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2 years ago