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IceJOKER [234]
3 years ago
6

If im pushing on a wall what are the action and reaction force?

Physics
1 answer:
artcher [175]3 years ago
5 0
When you push on a wall, the wall pushes back on you with a force equal in strength to the force you exerted.
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How does a virus reproduce
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<span>A virus particle is not a complete cell but an intracellular parasite. Hence, it cannot reproduce without the help of a host cell. Once inside a host cell, the virus is made in such a way that it replicates itself. There are two ways in which viruses reproduce or multiply their numbers: the lytic cycle and the lysogenic cycle. </span>
4 0
3 years ago
Read 2 more answers
about how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?
OverLord2011 [107]

Answer:

For example, an earthquake of magnitude 5.5 releases about 32 times as much energy as an earthquake measuring 4.5. Another way to look at this is that it takes about 900 magnitude 4.5 earthquakes to equal the energy released in a single 6.5 earthquake.

Explanation:

8 0
2 years ago
Determine the work that is being done by tension in pulling the box 198.0 cm along the table.
Julli [10]
Work, Kinetic Energy and Potential Energy
6.1 The Important Stuff 6.1.1 Kinetic Energy
For an object with mass m and speed v, the kinetic energy is defined as K = 1mv2
2
(6.1)
Kinetic energy is a scalar (it has magnitude but no direction); it is always a positive number; and it has SI units of kg · m2/s2. This new combination of the basic SI units is
known as the joule:
As we will see, the joule is also the unit of work W and potential energy U. Other energy
1joule = 1J = 1 kg·m2 (6.2) s2
units often seen are:
6.1.2 Work
1erg=1g·cm2 =10−7J 1eV=1.60×10−19J s2
When an object moves while a force is being exerted on it, then work is being done on the object by the force.
If an object moves through a displacement d while a constant force F is acting on it, the force does an amount of work equal to
W =F·d=Fdcosφ (6.3)
where φ is the angle between d and F.
5 0
3 years ago
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
4 years ago
Which of the following is a small star that has reached the end of its stellar evolution?
Akimi4 [234]
D.) White Dwarf

It is the smallest star whose mass is approximately equal or greater than 1.4M
Here, M = mass of the Sun.

Hope this helps!
3 0
3 years ago
Read 2 more answers
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