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3 years ago

15

I NEED HELP PLEASE, THANKS! :)

1 answer:

Luden [163]3 years ago

8 0

Answer:

The ball will be attracted to the negatively charged plate. It'll touch and pick up some electrons from the plate so that the ball becomes negatively charged. Immediately the ball is repelled by the negative plate and is attracted to the positive plate. The ball gives up electrons to the positive plate so that it is positively charged and suddenly attracts to the negative plate again, flies over to it and picks up enough electrons to be repulsed by negative plate and again to the positive plate and that continues.

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A house is wired so that one electrical source comes to a room but many outlets and lights work from that source . when one ligh

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The room is wired as a parallel circuit

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For most people, getting 8 to 9 hours of sleep increases concentration, improves physical health, and improves one's mood.

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3 years ago

Read 2 more answers

A 3250 N car is pushed a distance of 35 m the power was 11375 J, how long did it take?

irakobra [83]

Answer:

10.8s

Explanation:

Given parameters:

Force on the car = 3250N

Distance = 35m

Power = 11375W

Unknown:

Time taken = ?

Solution:

To solve this problem;

Power is the rate at which work is done

Power = $\frac{work done }{time}$

Work done = force x distance = 3250 x 35 = 123200J

Now;

11375 = $\frac{123200}{t}$

11375t = 123200

t = 10.8s

5 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg

nata0808 [166]

Answer:

a) $U_{e} = 3 \times 10^{10}\,J$ , b) $v \approx 7745.967\,\frac{m}{s}$

Explanation:

a) The potential energy is:

$U_{e} = Q \cdot \Delta V$

$U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)$

$U_{e} = 3 \times 10^{10}\,J$

b) Maximum final speed:

$U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }$

The final speed is:

$v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }$

$v \approx 7745.967\,\frac{m}{s}$

3 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago