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2 years ago

Select all that apply.

2 answers:

6 0

$\text{Acceleration}=\frac{d^2}{dt^2}\text{displacement}$
So the dimensions of acceleration is $LT^{-2}$
Any answer that comes under that definition is correct.

spin [16.1K]2 years ago

6 0

Strange as it may seem, four of those five choices
are perfectly good units of acceleration.

The only one that isn't is the one in the middle: ft/s

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The unit of electric current is the _____. A) volt B) ohm C) ampere D) faraday

professor190 [17]

the unit of electric current is the ampere, often shortened to amp

6 0

2 years ago

Read 2 more answers

A cheetah can run at approximately 100 km/hr and a gazelle at 80 km/hr. If both animals are running at full speed, with a gazell

zimovet [89]

The cheetah's speed is 100x and The gazelle's speed is 80x + 70. Set the two equations equal to each other: 100x = 80x +70 (then subtract 80x from both sides). 20x = 70 (then divide by 20). X =3.5. The cheetah catches the gazelle after 3.5

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3 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ????0 . Find the minimu

Reptile [31]

Answer:

Explanation:

Threshold frequency = 4.17 x 10¹⁴ Hz .

minimum energy required = hν where h is plank's constant and ν is frequency .

E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴

= 27.52 x 10⁻²⁰ J .

wavelength of radiation falling = 245 x 10⁻⁹ m

Energy of this radiation = hc / λ

c is velocity of light and λ is wavelength of radiation .

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹

= .08081 x 10⁻¹⁷ J

= 80.81 x 10⁻²⁰ J

kinetic energy of electrons ejected = energy of falling radiation - threshold energy

= 80.81 x 10⁻²⁰ - 27.52 x 10⁻²⁰

= 53.29 x 10⁻²⁰ J .

4 0

2 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

2 years ago

A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m

solmaris [256]

Answer:

The strength of the magnetic field that the line produces is $2x10^{-5} Tesla$ .

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

$B = \frac{\mu_{0}I}{2\pi r}$ (1)

Where $\mu_{0}$ is the permiability constant, I is the current and r is the distance from the wire.

Notice that it is necessary to express the current, I, from kiloampere to ampere.

$I = 1.0kA \cdot \frac{1000A}{1kA}$ ⇒ $1000A$

Finally, equation 1 can be used:

$B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}$

$B = 2x10^{-5}T$

Hence, the strength of the magnetic field that the line produces is $2x10^{-5} Tesla$ .

8 0

2 years ago