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2 years ago

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According to Newton's thirdlaw, every action has an equal and opposite reaction. When a comb is charged and held near small piec

es of paper, the comb exerts an electric force on the paper pieces and pulls them toward it. Why don't you observe the comb moving toward the paper pieces as well

1 answer:

NeTakaya2 years ago

5 0

Answer:

Due to the fact that the comb is in terms of mass bigger than pieces of paper

Explanation:

In terms of mass, the comb is way too bigger than the piece of paper so it is virtually impossible for it to move towards the pieces of paper

According to Newton third law action and reaction are equal and opposite

Mathematically

P=m*v

Where

P= impulse is Newton

m=mass of the body

V= velocity of the body

Now since the comb has a greater mass it will not move to the paper, else the paper must have to move to it.

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An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt

rusak2 [61]

If the velocity is constant then the acceleration of the object is zero.
$a=0 (m/s^2)$
Thus when we apply the equation
$\Delta X=vt+(at^2/2)$
It remains
$\Delta X =vt$
or equivalent
$t=(\Delta X/v) =7500/278 =26.98 (seconds)$

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3 years ago

An input for of 80 N is used to lift an object weighing 240 N with a system of pulleys. How far must the rope around the pulleys

sammy [17]

Answer:

4.2 m

Explanation:

Note: If energy is conserved, i.e no work is done against friction

Work input = work output.

Work output = Force output × distance,

Work input = force input × distance moved moved.

Therefore,

input force×distance moved = output force × distance moved........................Equation 1

Given: input force = 80 N, output force = 240 N, output distance = 1.4 m

Let input distance = d

Substitute into equation 1

80×d = 240×1.4

80d = 336

d = 336/80

d = 4.2 m.

Thus the rope around the pulley must be pulled 4.2 m

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Explanation:

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3 years ago

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An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in

weqwewe [10]

Answer:

a) $t=1.4\times 10^{-5}\ s$

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

a)

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

$t=1.4\times 10^{-5}\ s$

b)

We know that

F = m a = E q ------------1

Mass of electron ,m

$m=9.1\times 10^{-31}\ kg$

Charge on electron

$q=1.6\times 10^{-19}\ C$

So now by putting the values in equation 1

$a=\dfrac{E q}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2$

$a=4.74\times 10^{9}\ m/s^2$

$S= ut+\dfrac{1}{2}at^2$

Here initial velocity u= 0 m/s

$S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m$

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

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3 years ago

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Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

nalin [4]

Answer:

$13 W/m^2$

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

$I \propto \frac{1}{r^2}$

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

$\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2}$ (1)

where in this problem, we have:

$I_1 = 1300 W/m^2$ apparent brightness at a distance $r_1$ , where

$r_1 = 150$ million km

We want to estimate the apparent brightness at $r_2$ , where $r_2$ is ten times $r_1$ , so

$r_2 = 10 r_1$

Re-arranging eq.(1), we find $I_2$ :

$I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2$

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2 years ago