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2 years ago

14

If a neutral object such as paper comes close to a positively charged plastic rod, what type of charge accumulates on the side o

f the paper closest to the positive rod?

2 answers:

Tju [1.3M]2 years ago

7 0

Explanation:

When a neutral object such as paper comes close to a positively charged plastic rod then an opposite charge, that is, negative charge will develop on the paper.

As a result, the side of paper with negative charge will get attracted towards the positive charge of plastic rod. As it is known that opposite charges attract each other and like charges repel each other.

Nataliya [291]2 years ago

3 0

The answer would be negative charge because +, and - dont like each other so they retract from each other.

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

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2 years ago

Dentists' chairs are examples of hydraulic-lift systems. If a chair weighs 1400 N and rests on a piston with a cross-sectional a

NeX [460]

Answer:

Force applied to smaller cross section is

= 82.63 N

Explanation:

As we know

$F_2 A_1 = F_1 A_2$

where $F1, F2$ signifies the weight of the two chair in a hydraulic-lift system

And $A_1, A_2$ signifies the area of the two respective chairs in a hydraulic-lift system

Given -

$F2=1400$ N

$A1 =1220$ Square centimeter

$A_2 = 72$ Square centimeter

Substituting the given values in above equation, we get -

$1400 * 72 = F1 * 1220\\F2 = 82.63$

Force applied to smaller cross section is

= 82.63 N

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2 years ago

A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the

likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

$a\sin \theta_n=n\lambda$

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

$a\sin \theta_1=(1)\lambda$

solving for θ1:

$\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_1=0.63 degrees$

for the second minimum:

$a\sin \theta_2=(2)\lambda$

$\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_2=1.26 degrees$

So, the angular separation between them is the rest:

$\Delta \theta =1.26-0.63$

$\Delta \theta=0.63$

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2 years ago

The hammer throw was one of the earliest Olympic events. In this event, a heavy ball attached to a chain is swung several times

Aleonysh [2.5K]

Answer:

Given that

T= 0.43 s

Radius of the ball path's , r=2.1 m

a)

We know that

f= 1/T

Here f= frequency

T= Time period

Now by putting the values

f= 1/T

T= 0.43 s

f= 1/0.43

f=2.32 Hz

b)

We know that

V= ω r

ω = 2 π f

ω=Angular speed

V= Linear speed

ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s

V= ω r= 14.60 x 2.1 = 30.66 m/s

c)

Acceleration ,a

a =ω ² r

a= 14.6 ² x 2.1 = 447.63 m/s²

We know that g = 10 m/s²

So

a= a/g= 447.63/10 = 44.7 g m/s²

a= 44.7 g m/s²

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2 years ago

A student placed a pencil in a cup of water. The pencil appears broken because light- always travels in a straight line makes th

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<span>None of the choices makes a correct statement. The third choice is close,
but misleading.

The pencil appears broken because light bends away from a straight line
when it crosses the boundary between air and water.</span>

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