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3 years ago

What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?

1 answer:

7 0

We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
$\Delta V = I R$
where
$\Delta V$ is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and $R=330 \omega$ , so the potential difference is
$\Delta V = IR=(4.00 A)(330 \omega)=1320 V$

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Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

Gravity at a height h above the surface of the moon will be given as:

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

2 years ago

One baked potato provides an average of 30 mg of vitamin C. If 70 potatoes weigh 20 lb., how many milligrams of vitamin C are pr

JulsSmile [24]

Answer:

105 mg

Explanation:

Given that:

1 baked potato provides 30 mg of vitamin C.

So,

70 baked potatoes provide $30\times 70$ mg of vitamin C

Also,

70 potatoes = 20 lb

So,

20 lb potatoes provide $30\times 70$ mg of vitamin C

Thus,

1 lb potatoes provide $\frac {30\times 70}{20}$ mg of vitamin C

Thus, 105 mg of Vitamin C are provided per pound of the potatoes.

5 0

3 years ago

The density, or intensity, of a magnetic field is called flux. True False

aev [14]

Answer:

True :)

Explanation:

4 0

2 years ago

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posledela

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3 years ago

Read 2 more answers

The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

pav-90 [236]

A) $5.0\cdot 10^{-11} m$

The energy of an x-ray photon used for single dental x-rays is

$E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J$

The energy of a photon is related to its wavelength by the equation

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

Re-arranging the equation for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m$

B) $2.0\cdot 10^{-11} m$

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

$E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J$

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m$

4 0

2 years ago