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A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti

matrenka [14]

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

$\phi = \frac{Q}{\epsilon_{o}}$ (Gauss's Law)

Putting the given values into the above formula as follows.

Q = $\phi \times \epsilon_{o}$

= $-953 Nm^{2}/C \times 8.854 \times 10^{-12}$

= $-8.4 \times 10^{-9} C$

= -8.4 nC

Therefore, when the unknown charge is q then,

-14.0 nC + 33.0 nC + q = -8.4 nC

q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

7 0

4 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

4 0

3 years ago

Which three factors are used to calculate gravitational potential energy?

Elis [28]

Explanation:

PEgrav = m *• g • h

In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.

www.physicsclassroom.com › energy

Potential Energy - The

8 0

4 years ago

Why are the element from period 2 grouped together

valentinak56 [21]

All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons. All of the elements in the second row (the second period) have two orbitals for their electrons. As you move down the table, every row adds an orbital.

4 0

4 years ago

A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface. The surface of the ramp is the x-

kifflom [539]

Magnitude of the force of tension: 139 N

Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

The force of tension, $F_T$ , acting up along the plane
The force of friction, $F_f=14.8 N$ , acting down along the plane
The component of the weight in the x-direction, $F_{gx}$ , acting down along the plane