What are the differences between refraction and diffraction?

The potential energy of an object attached to a spring is 2.60 J at a location where the kinetic energy is 1.40 J. If the amplit

anyanavicka [17]

Answer:

This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.

U = (1/2)kx^2

U = (1/2)(5.3)(3.62-2.60)^2

U = 2.75706 J

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Explanation:

6 0

3 years ago

What scientific question does Isaac Asimov most clearly address with his

natali 33 [55]

Answer: C. can robots be trained to follow a moral code?

Explanation:

6 0

3 years ago

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What happens when a roller coaster car moves down from the top of a hill?

VLD [36.1K]

When the roller coaster moves down from the top of the hill, all of its stored potential energy is converted into kinetic energy to move it and when it goes back up the hill it turns kinetic into potential.

3 0

3 years ago

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An AC generator supplies 230 V peak voltage at 210 Hz frequency. How much power is dissipated in a resistor with R = 225 Ω resis

sesenic [268]

Answer:

The power dissipated in a resistor is 117.54 watts.

Explanation:

Given that,

Peak voltage of the Ac generator, V = 230 V

Frequency, f = 210 Hz

Resistance, R = 225 ohms

We need to find the power dissipated in a resistor. The power generated is given by :

$P=\dfrac{V^2_{rms}}{R}$

$V_{rms}=\dfrac{V}{\sqrt{2} }\\\\V_{rms}=\dfrac{230}{\sqrt{2} } \\\\V_{rms}=162.63\ V$

So,

$P=\dfrac{162.63^2}{225}\\\\P=117.54\ W$

So, the power dissipated in a resistor is 117.54 watts. Hence, this is the required solution.

7 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago