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cluponka [151]
3 years ago
7

What are the laws of Physics​

Physics
1 answer:
uranmaximum [27]3 years ago
8 0

Answer:

thermodynamics

Explanation:

The laws of thermodynamics define a group of physical quantities, such as temperature, energy, and entropy, that characterize thermodynamic systems in thermodynamic equilibrium.

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Which term describes the tendency of an organism to regulate internal conditions for maintaining good health?
meriva
Homeostasis: this is the body’s way of creating an internal state of equilibrium.
3 0
3 years ago
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olganol [36]

Answer:

I Dont know u answer it

Explanation:

8 0
2 years ago
Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia
Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

<u>Wn = 9.14 x 10¹⁷ N</u>

3 0
3 years ago
What is induced current​
Brrunno [24]

Answer:

A current can be induced in a conducting loop if it is exposed to a changing magnetic field. ... In other words, if the applied magnetic field is increasing, the current in the wire will flow in such a way that the magnetic field that it generates around the wire will decrease the applied magnetic field.

Explanation:

8 0
3 years ago
A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o
Nataly_w [17]

Explanation:

The given data is as follows.

      C_{1} = 1.50 \times 10^{-6} F

      C_{1} = 3.50 \times 10^{-6} F    

      Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

    \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

    \frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}

                 = 0.945 \times 10^{-6} F

          C = 1.06 \times 10^{-6} F

Now, we will calculate the charge across each capacitance as follows.

              Q = CV

                  = 1.06 \times 10^{-6} F \times 2.50 V

                  = 2.65 \times 10^{-6} C

                  = 2.65 \mu C

Thus, we can conclude that 2.65 \mu C is the charge stored on each given capacitor.

5 0
3 years ago
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