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3 years ago

A 5.0-C charge is 10m from a -2.0-C charge. The electrostatic force on the positive charge is

Physics

2 answers:

r-ruslan [8.4K]3 years ago

8 0

Answer:

The electrostatic force away from the negative charge is $9\times10^{8}\ N$

(b) is correct option.

Explanation:

Given that,

First charge = 5.0 C

Second charge = -2.0 C

Distance = 10 m

We need to calculate the electrostatic force on the positive charge

Using formula of electrostatic force

$F=\dfrac{kq_{1}q_{2}}{r^2}$

Where, q₁ = first charge

q₂= second charge

r = distance

Put the value into the formula

$F_{12}=\dfrac{9\times10^{9}\times5.0\times(-2.0)}{10^2}$

$F_{12}=−900000000\ N$

$F_{12}=-9\times10^{8}\ N$

The electrostatic force away from the negative charge.

Hence, The electrostatic force away from the negative charge is $9\times10^{8}\ N$

VMariaS [17]3 years ago

7 0

Answer:

9.0 × 108 N toward the negative charge

Explanation:

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3 years ago

Read 2 more answers

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w

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Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

10 x 3 + 25 x 0 = v(10 + 25)

30 = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

$K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\$

the final kinetic energy of the package;

$K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J$

the fraction of the initial energy lost;

$= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71$

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3 years ago

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3 years ago

A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel

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Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

$T= 2\pi\sqrt{\frac{M}{k}}$

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

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3 years ago

A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i

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Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

Solution:

As per the question:

The eqn of the displacement is given by:

$y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]$ (1)

n = 4

Now,

We know the standard eqn is given by:

$y = AsinKxcos\omega t$ (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = $14.4 m^{- 1}$

$\omega = 166\ rad/s$

where

A = Amplitude

K = Propagation constant

$\omega$ = angular velocity

Now, to calculate the string's wavelength,

(a) $K = \frac{2\pi}{\lambda}$

where

K = propagation vector

$\lambda = \frac{2\pi}{K}$

$\lambda = \frac{2\pi}{14.4} = 0.436\ m$

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$l = \frac{n\lambda}{2}$

$l = \frac{4\times 0.436}{2} = 0.872\ m$

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$f = \frac{\omega}{2\pi}$

$f = \frac{2\pi}{166} = 26.42\ Hz$

Now,

Speed of the wave is given by:

$v = f\lambda$

$v = 26.419\times 0.436 = 11.518\ m/s$

4 0

3 years ago