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2 years ago

What is a living organism? Give a few examples.

Physics

1 answer:

zhuklara [117]2 years ago

7 0

Answer:

It's a individual form of life. Examples of this are bacterium , protists and fungus

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PLEASEEE HELPPPPPPP:

tatuchka [14]

V(voltage) = I(current)R(resistance)
substitute in the values

V = 15 * 0.10
V = 1.5 volts

7 0

3 years ago

Read 2 more answers

How do I solve for the maximum speed and height given those accelerations? (please give the formula so I can solve these types o

Sphinxa [80]

It depends on how you want to solve it you can solve it in many different meathods:$

5 0

2 years ago

Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu

Elenna [48]

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)² = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424 x 10²⁵

= 6.4 x 10²³ kg .

4 0

2 years ago

If a series circuit contains a 12-V battery, a 6-ohm resistor, and a 4-ohm resistor, what is the current in the circuit?

Shalnov [3]

In a series circuit the total current is the same throughout resistors and so:

$I_{total}=I_1=I_2$

The voltage is distributed throughout the resistors and so:

$V_{total}=V_1+V_2$

and the total resistance can be calculated by adding up the resistors resistance:

$R_{total}=R_1+R_2$

First thing is to calculate the total resistance and so:

$R_{total}=6\Omega + 4\Omega = 10\Omega$

And by Omh's law V=IR we have:

$V_{total}=I_{total}R_{total}\\\\I_{total}=\frac{V_{total}}{R_{total}}= \frac{12V}{10\Omega} =1.2A$

And so the total current of the circuit is 1.2 amps i.e. 1.2 A.

6 0

3 years ago

Read 2 more answers

Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k

Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

$\frac{GM_{ns}}{R^2}= \omega^2 R$

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

$M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}$

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0

3 years ago