Ask question

Ask question

All categories

3 years ago

6

How far apart (in mm) must two point charges of 70.0 nC (typical of static electricity) be to have a force of 1.30 N between the

m?

1 answer:

Sever21 [200]3 years ago

5 0

Answer:r=5.824 mm

Explanation:

Given

Charge $q_1=70 nC$

$q_2=70 nC$

Force between them $F=1.30 N$

Electrostatic Force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ are the charge on the Particles

r=distance between them

$k=Coulomb\ constant =9\times 10^9 N-m^2/C^2$

$1.3=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{r^2}$

$r^2=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{1.3}$

$r=\sqrt{33.923\times 10^{-6}}$

$r=5.824\times 10^{-3}$

$r=5.824 mm$

You might be interested in

Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up

Westkost [7]

Answer:

24 N

Explanation:

$m$ = mass of the cube = $6.0 kg$

Consider the three cubes together as one.

$M$ = mass of the three cubes together = $3 m = 3 (6.0) = 18 kg$

$a$ = acceleration of the combination = 2 ms⁻²

$F$ = Force applied on the combination

Using Newton's second law

$F = ma = (18) (2) = 36 N$

$F_{L}$ = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

$F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N$

4 0

3 years ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

3 years ago

7.1 Project Guidelines 2021

Ede4ka [16]

I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and

3 0

2 years ago

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep

Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

m1 = 0.0225 Kg

V1 = (0.869, -0.283) m / s

Px = m Vx

Px1 = 0.0225 0.869

Px1 = 0.01955 Kg m

Py = m Vy

Py1 = 0.0225 (-0.283)

Py1 = -0.006368 Kg m

P1 = (0.0196, -0.00637) Kg m

Mouse 2

m2 = 0.0223 Kg

Px2 = 0.0223 (-0.883) = -0.0196 Kg m

Py2 = 0.0223 (-0.253) = -0.00564 Kg m

P2 = (-0.0196, -0.00564) Kg m

Mouse 3

m3 = 0.0197

Px3 = 0.0197 0.345 = 0.00680 Kg m

Py3 = 0.0197 0.803 = 0.0158 Kg m

P3 = (0.00680, 0.0158) Kg m

Mouse4

m4 = 0.0127 Kg

Px4 = 0.0127 (-0.555) = -0.00705 Kg m

Py4 = 0.0127 0.205 = 0.00260 Kg m

P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

Px = Px1 + Px2 + Px3 + Px4

Py = py1 + Py2 + Py3 + Py4

Px = 0.0196 -0.0196 +0.00680 -0.00705

Px = -0,000250 Kg m

Py = -0.00637 -0.00564 +0.0158 +0.00260

Py = 0.00639 Kg m

P = (-0.000250, 0.00639) Kg m

7 0

3 years ago

Which of these is NOT an example of balanced forces? A. lying still on a bed B. a ship slowly sinking C. leaning against a brick

liraira [26]

<span>B. a ship slowly sinking</span>
This is not balanced

4 0

3 years ago

Read 2 more answers