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3 years ago

ELE Review questions

Physics

1 answer:

zheka24 [161]3 years ago

4 0

Answer:

1. ELE to KE = Electrical energy conversion to kinetic energy.

2. KE to ELE = kinetic energy conversion to electrical energy.

3. CE to ELE = chemical energy conversion to electrical energy.

Explanation:

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

Some examples of energy are chemical energy, electrical energy, kinetic energy, sound energy, potential energy, light energy, etc.

The transformation of energy from one form to another are described below;

1. A hair dryer plugged into the wall: it involves the conversion of electrical energy to kinetic energy. The energy type present in the wall socket is electrical energy which powers the hair dryer to have kinetic energy i.e the energy possessed due to motion of hair dryer.

2. A hand crank generator emergency radio: it illustrates the conversion of kinetic energy to electrical energy. The hand crank generator possess kinetic energy as it is being wounded before it is then converted into electrical energy which powers the radio.

3. A battery operated fan: it illustrates the conversion of chemical energy to electrical energy. The cells present in battery possess chemical energy that is being converted into electrical energy to power the fan.

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A helicopter flies 168 miles against the wind in 2 hours; with the wind, it can fly 252 miles in the same amount of time. find t

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Let $v_h$ be the speed of the helicopter in still air. Let $v_w$ be the speed of the wind. Then, from the given information,

$(v_h-v_w)2=168\\ (v_h+v_w)2=252\\$

Adding the above 2 equations,

$4v_h=168+252\\ 4v_h=420\\ v_h=105$

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What is a good example of a contact force?

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4 years ago

QuestionDetails:

Alexxandr [17]

Answer:

The sketch for the Gravitational force F and the potential energy U are attached to this answer.

Explanation:

To obtain the gravitational force, we can consider the gravitational field GF(r) as:

$\vec{F_{g}}=m_{par} \cdot \vec{G_F(r)}$

To calculate the gravitational field we can use the Gauss theorem. By considering a homogeneous mass of the sphere (constant density) and the spherical symmetry, we can determinate than the gravitational field direction is $-\hat{r}$ .

Considering a constant density:

$Vol=\pi\frac{4}{3}(a^3-b^3)$

$\rho=\frac{m_{sph}}{Vol} =\frac{3m_{sph}}{4\pi(a^3-b^3)}$

Applying a spherical gaussian surface for different radius r:

for R<b:

$\displaystyle\oint_{S} \vec{G_{f}}\, \vec{ds}=-4m_{int}G=0N/kg \rightarrow G_{f}=0N/kg \rightarrow F_{g}=0N$

for b<R<a:

$\displaystyle\oint_{S} \vec{G_{f}}\, \vec{ds}=-4\pi m_{int}G\\\displaystyle\oint_{S} -G_{f}\, ds=-4\pi G\displaystyle\int_{} \int_{} \int_{V} \rho\, dV\,\,\\\\-4\pi r^2 G_{f}(r)=-4\pi Gm_{sph}\frac{(r^3-b^3)}{(a^3-b^3)} \\G_{f}(r)=Gm_{sph}\frac{(r^3-b^3)}{r^2(a^3-b^3)} \\F_g =Gm_{par}m_{sph}\frac{(r^3-b^3)}{r^2(a^3-b^3)}(-\hat{r})$

for a<R:

$\displaystyle\oint_{S} \vec{G_{f}}\, \vec{ds}=-4\pi m_{int}G\\\\-4\pi r^2 G_{f}(r)=-4\pi Gm_{sph} \\ F_g =Gm_{par}m_{sph}\frac{1}{r^2}(-\hat{r})$

For the potential energy you can integrate the field to obtain the gravitational potential and the multiplying for the particle mass:

for a<R:

$U(r)=G\frac{m_{sph}m_{par}}{r}$

for b≤R≤a:

$U(r)=G\frac{m_{sph}m_{par}}{a} +G\frac{m_{sph}m_{par}}{(a^3-b^3)} (\frac{a^{2}}{2}+\frac{b^{2}}{a}-\frac{r^{2}}{2}-\frac{b^{2}}{r})$

for R≤b:

$U(r)=G\frac{m_{sph}m_{par}}{a} +G\frac{m_{sph}m_{par}}{(a^3-b^3)} (\frac{a^{2}}{2}+\frac{b^{2}}{a}-\frac{b^{2}}{2}-b)$

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