If you lift a five pound object 18 inches how many joules of energy did yo use?

A 3.6-volt battery is used to operate a cell phone

Marina CMI [18]

Explanation :

It is given that,

Potential energy, $V=3.6\ V$

Power dissipated, $P=0.064\ Watt$

We know that the power dissipated is given by :

$P=VI$

I is the current passing through the phone.

$I=\dfrac{P}{V}$

$I=\dfrac{0.064\ W}{3.6\ V}$

$I=0.017\ A$

or

I = 0.018 A

Hence, the current that passes through the phone is (1) 0.018 A.

3 0

3 years ago

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In comparison with a steep slope, a gentle slope _____.

kifflom [539]

A gentle slope requires less force over a longer distance as compared to steep slope.

Explanation:

Mechanical advantage of a slope is equal to the ratio of length of slope and the height. A steep slope has shorter length as compared to a gentle slope for the same height. Therefore, mechanical advantage of a gentle slope is more than that of a steep slope. Hence, a gentle slope requires less force over a long distance than a steep slope.

6 0

3 years ago

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A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

grin007 [14]

Answer:

$f_e = 1.51 cm$

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

$m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}$

$400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}$

$9.6 = \dfrac{16-f_e}{f_e}$

$9.6f_e = 16-f_e$

$10.6 f_e = 16$

$f_e = 1.51 cm$

3 0

3 years ago

What us the difference between accurate data and reproducible data

Nikolay [14]

It's just in the name! Accurate data is helpful, and correct, but reproducible data is all of that, and is able to be given to other people through different sources! At least, that's what my understanding of them are. Hope it helps!

4 0

3 years ago

An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi

elena55 [62]

Answer:

a) $\vec F_{B} = 8.766\times 10^{-14}\,T\,k$ , b) $\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

$\vec F_{B} = q\cdot \vec v \times \vec B$

The charge of the electron is equal to $-1.602\times 10^{-19}\,C$ . Then, cross product can be solved by using determinants:

$\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = 8.766\times 10^{-14}\,T\,k$

b) The charge of the proton is equal to $1.602\times 10^{-19}\,C$ . Then, cross product has the following determinant:

$\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

8 0

2 years ago