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3 years ago

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A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a

uniformly distributed charge of -7 nC (see the figure) on its outer surface. The distance between the centers of the balls is 9 cm.Assume the +x axis runs to the right, the +y axis runs up and the +z axis is out.What is the electric field at the center of the metal ball due only to the charges on the surface of the metal ball?

1 answer:

blsea [12.9K]3 years ago

4 0

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

$E=7*10^{9}N/C$

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field is mathematically given as

$E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\$

$E=7*10^{9}N/C$

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A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T

den301095 [7]

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

$m_pv_p=v_c(m_p+m_b)$ where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

$0.5(m_p+m_b)v_c^{2}=0.5kx^{2}$ where k is spring constant and x is the compression of spring. Substituting the given values then

$(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m$

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3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?

<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>

<h3>Given:-</h3>

Velocity (V) = 40 m/s

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