All categories

2 years ago

The amount of work required to increase the distance between -6μC and 4μC from 6 cm to 18 cm will be

Physics

1 answer:

White raven [17]2 years ago

4 0

Answer:

W = 1.8 J

Explanation:

The amount of work required to move the given charges can be found by using the following formula:

$W = \frac{kq_1q_2}{\Delta r} \\\\$

where,

W = Work done = ?

k = Colomb's constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 6 μC = 6 x 10⁻⁶ C

q₂ = magnitude of second charge = 4 μC = 4 x 10⁻⁶ C

Δr = change in distance = 18 cm - 6 cm = 12 cm = 0.12 m

Therefore,

$W = \frac{(9\ x\ 10^9\ Nm^2/C^2)(6\ x\ 10^{-6}\ C)(4\ x\ 10^{-6}\ C)}{0.12\ m}$

You might be interested in

The mcb of rupa's room is tripped and keeps on tripping again and again . if it is a domestic circuit, what could be the reason

Tamiku [17]

The MCB of a rupas room is tripped and keeps on tripping again and again, and if it is a domestic circuit, what could be the reason for this phenomenon?

The reason could be a short circuit which is resulting in higher level of currents to pass through the MCB which is resulting in trip every time.

The MCB is faulty and might need a replacement.

To Diagnose the problem further more.

Turn off all the switches in rupas room and then try turning on the MCB. If it trips again then MCB is faulty (Subjective to the fact there everything was normal before this issue and no signs of short circuit or spark in wiring were observed)

If MCB does not trip in point 1 then Turn ON all the switches one by one. This shall give you the cause of problem.

4 0

2 years ago

When an objects total mechanical energy is conserved and it is dropped from rest how does the objects initial potential energy c

frozen [14]

Answer:Magnitude of potential energy before falling will be equal to the magnitude of the final kinetic energy, energy is only converted from potential to kinetic

Explanation: law of conservation of energy says that energy can neither be created or destroyed but Change from one form to another.so nothing will happen to the size of the potential energy when it changes to kinetic energy

8 0

2 years ago

3. A model rocket is launched straight upward at 58.8 m/s.

SIZIF [17.4K]

Let us assume that rocket only runs in initial energy and not using its own to flying.

Also , let upward direction is +ve and downward direction is -ve .

Initial velocity , u = 58.8 m/s .

Acceleration due to gravity , $g=-9.8\ m/s^2$ .

Final velocity , v - = 0 m/s .

We know , by equation of motion .

$v^2-u^2=2gh\\\\2gh_{max}=0^2-58.8^2\\\\h_{max}=\dfrac{0^2-58.8^2}{-2\times 9.8}\\\\h_{max}= 176.4\ m$

Hence, this is the required solution .

8 0

2 years ago

The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P

lesya [120]

Answer:

a) For P: $v=0.938\frac{m}{s}$

For Q: $v = 1.876\frac{m}{s}$

b) For P:

$a_{rad}=8.80\frac{m}{s^{2}}$

for Q:

$a_{rad}=17.60\frac{m}{s^{2}}$

c) As the distance from the axis increases then speed increases too.

Explanation:

a) Assuming constant angular acceleration we can find the angular speed of the wheel dividing the angular displacement θ between time of rotation:

$\omega =\frac{\theta}{t}$

One rotation is 360 degrees or 2π radians, so θ=2π

$\omega =\frac{2\pi}{0.670} =9.38\frac{rad}{s}$

Angular acceleration is at every point on the wheel, but speed (tangential speed) is different and depends on the position (R) respect the rotation axis, the equation that relates angular speed and speed is:

$v = \omega R$