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Sladkaya [172]
3 years ago
8

the body Will remain at rest or move at constant velocity unless acted by external net force are unbalance force​

Physics
1 answer:
Vikentia [17]3 years ago
4 0

Answer:

law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia. Inertia is the tendency of an object to remain at rest or remain in motion.

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The acceleration of an object is proportional to which of the following
Lana71 [14]

the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

7 0
3 years ago
true or false? Atoms that have the same number of neutrons but different numbers of protons are called isotopes. WHY?
-Dominant- [34]
It is false because it is the reverse of it
6 0
4 years ago
Read 2 more answers
Gaseous helium is in thermal equilibrium with liquid helium at 6.4 K. The mass of a helium atom is 6.65 × 10−27 kg and Boltzmann
chubhunter [2.5K]

Answer:

162.78 m/s is the most probable speed of a helium atom.

Explanation:

The most probable speed:

v_{mp}=\sqrt{\frac{2K_bT}{m}}

K_b= Boltzmann’s constant =1.38066\times 10^{-23} J/K

T = temperature of the gas

m = mass of the gas particle.

Given, m = 6.65\times 10^{-27} kg

T = 6.4 K

Substituting all the given values :

v_{mp}=\sqrt{\frac{2\times 1.38066\times 10^{-23} J/K\times 6.4 K}{6.65\times 10^{-27} kg}}

v_{mp}=162.78 m/s

162.78 m/s is the most probable speed of a helium atom.

4 0
4 years ago
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

8 0
4 years ago
Which of these black holes exerts the weakest tidal forces on something near its event horizon? a. a 10Msun black hole b. a 100M
Darya [45]

The answer is A. You are welcome.

7 0
4 years ago
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