(10p+15,15-10q)=(25,5)

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the

marin [14]

Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$x$ = initial position of spring from equilibrium position = 0.21 m

$v_{i}$ = initial speed of the cart = 2.0 ms⁻¹

$A$ = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

$(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m$

B)

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$A$ = amplitude of the oscillation = 0.395 m

$v_{o}$ = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

$(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}$

5 0

3 years ago

Read 2 more answers

The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b

Romashka-Z-Leto [24]

Answer:

I=0.0361 kg.m^2

Explanation:

Torque is the rotational equivalent of a force

Torque= perpendicular distance r X Force F

Torque T = I(moment of inertia) X α (angular acceleration)

T= Iα

r= 0.0285m

F= 1.9 x 10^3

T=0.0285 x 1.9 x 10^3

T= 54.15Nm

I=T/α

I=54.15/150

I=0.361 kg.m^2

4 0

2 years ago

find the x-component of this vector 18.4,0.250. Remember, angles are measured from the +x axis. Find the x-component and y-compo

Ksivusya [100]

Answer:

x = 0.237

y = 0.0789

Explanation:

Vector with direction 18.4° and magnitude 0.250 has x and y components of:

x = 0.250 cos 18.4°

x = 0.237

y = 0.250 sin 18.4°

y = 0.0789

7 0

2 years ago

Read 2 more answers

When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi

ryzh [129]

Answer:

<u>The pendulum bob swing past the mean position because:</u>

When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to the restoring toque the bob starts to accelerates towards its mean postion. The maximum acceleration of the pendulum's bob is $-w^{2} A$ and the the acceleration decreases as $-w^{2} x$ towards the mean position.

The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.

6 0

2 years ago

(10p+15,15-10q)=(25,5)​

(10p+15,15-10q)=(25,5)