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Anestetic [448]

3 years ago

5

To make the jump, Neo and Morpheus have pushed against their respective launch points with their legs applying a _____ to the la

unch points. distance dimension equilibrium force

1 answer:

ra1l [238]3 years ago

7 0

Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction

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A 2 N and an 8 N force pull on an object to the right and a 4 N force pulls on the same object to the left. If the object has a

Elden [556K]

Answer:

a = 12 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

where:

ΣF = sum of forces acting on a body [N] (units of Newtons)

m = mass = 0.5 [kg]

a = acceleration [m/s²]

Let's take the direction of positive forces to the right and negative forces directed to the left

2 + 8 - 4 = 0.5*a

6 = 0.5*a

a = 12 [m/s²]

7 0

2 years ago

Which of the following statements best supports the idea that science strives for objectivity?

nasty-shy [4]

I am going to have to say B

6 0

3 years ago

Read 2 more answers

Is the SI unit of work newton?

Naya [18.7K]

Answer:

Newton is the SI unit for force . Newton is kg m2

7 0

3 years ago

Read 2 more answers

Two bullets of the same size, mass and horizontal velocity are fired at identical blocks, only one is made of steel and the othe

tatiyna

Answer and Explanation:

Since we're discussing shots, the significant thing is the way the energy is changed over as there is deceleration of the bullet to a halt when it hits something.

Kinetic Energy is relative to mass times speed squared, so in reality, the 2 cases given have practically indistinguishable Kinetic energy. The measure of energy is authoritative, so the two cases will do generally a similar harm given, obviously we look at situations when all the kinetic energy is spent.

One contrast that will be effectively obvious is that the weapon in the case of heavy bullet will recoil more.
One can consider energy assimilation as force times separation distance, and energy ingestion as a product of force and time.
Henceforth, the heavier yet more slow bullet with a similar energy will venture to every part of a similar separation in the engrossing material, but since of bigger force, will take a more drawn out time doing it.
It will along these lines, additionally, give a more noteworthy "kick" to the object that absorbs.

8 0

3 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

3 years ago