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3 years ago

The topics covered by the study of thermodynamics include

1 answer:

5 0

A. energy changes during heating

Thermodynamics is the study of heat, "thermo," and work, "dynamics." ... Concepts covered in this tutorial include the laws of thermodynamics, internal energy, heat, work, PV diagrams, enthalpy, Hess's law, entropy, and Gibbs free energy.

please vote brainliest if this helped and was correct. :)

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Which of these types of images is not possible with a diverging mirror? A) an image that is smaller than the object b)a inverted

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Taking a shot in the dark here as i mostly deal with atomic structure and the fact i haven't handled a diverging mirror ut would it be a virtual image because that needs parallel normal lines which cannot occur

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3 years ago

Which of the following is most useful to determine how much energy is being used by a circuit in a given amount of time?

user100 [1]

Answer:

The answer is A.

Explanation:

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3 years ago

A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −

romanna [79]

Answer:

The speed of electron is $1.5\times10^{7}\ m/s$

Explanation:

Given that,

Electric field $E=1.50\times10^{3}\ N/C$

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy $P.E=5.04\times10^{-17}\ J$

Change in potential energy $\Delta P.E=-9.60\times10^{-17}\ J$ as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

$W=-eE\Delta x$

Put the value into the formula

$W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))$

$W=-5.04\times10^{-17}\ J$

We need to calculate the initial velocity

Using change in kinetic energy,

$\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2$

$\Delta K.E=\dfrac{-3mv^2}{8}$

Now, using work energy theorem

$\Delta K.E=W$

$\Delta K.E=\Delta U$

So, $\Delta U=W$

Put the value in the equation

$\dfrac{-3mv^2}{8}=-5.04\times10^{-17}$

$v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}$

Put the value of m

$v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}$

$v=1.21\times10^{7}\ m/s$

We need to calculate the change in potential energy

Using given potential energy

$\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})$

$\Delta U=-4.56\times10^{-17}\ J$

We need to calculate the speed of electron

Using change in energy

$\Delta U=-W=-\Delta K.E$

$\Delta K.E=\Delta U$

$\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}$

Put the value into the formula

$v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}$

$v_{f}=1.5\times10^{7}\ m/s$

Hence, The speed of electron is $1.5\times10^{7}\ m/s$

4 0

3 years ago

NEED ANSWER NOW PLZ!!

ira [324]

Answer:

Explanation:

A beta particle is formed when a neutron changes into a proton and a high-energy electron. The proton stays in the nucleus but the electron leaves the atom as a beta particle.

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3 years ago