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Anika [276]
3 years ago
14

Explain the Impulse-Momentum Theorem.

Physics
1 answer:
Inessa05 [86]3 years ago
4 0

The impulse-theorem states that the change in momentum of an object is equal to the impulse exerted on it

Explanation:

The impulse-theorem states that the change in momentum of an object is equal to the impulse exerted on it.

Mathematically, we have:

- The impulse is defined as the product between the force exerted (F) and the duration of the collision (\Delta t):

I=F \Delta t

- The change in momentum is equal to the product between the mass of the object (m) and the change in velocity (\Delta v):

\Delta p = m \Delta v

So, the theorem can be written as

F\Delta t = m \Delta v

This theorem can be proved by using Newton's second law. In fact, we know that

F=ma (1)

where a is the acceleration of the object. However, we can re-write the acceleration as the rate of change of velocity:

a=\frac{\Delta v}{\Delta t}

Therefore, (1) becomes:

F=m\frac{\Delta v}{\Delta t}

And by re-arranging,

F\Delta t = m \Delta v

Which is exactly the formula of the impulse theorem.

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

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Ira Lisetskai [31]

Answer:

0 < r < r_exterior     B_total = \frac{\mu_o I}{2\pi  r}

r > r_exterior            B_total = 0

Explanation:

The magnetic field created by the wire can be found using Ampere's law

        ∫ B. ds = μ₀ I

bold indicates vectors and the current is inside the selected path

           

outside the inner cable

          B₁ (2π r) = μ₀ I

          B₁ = \frac{\mu_o I}{2\pi  r}

the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.

For the outer shell

for the case   r> r_exterior

         

           B₂ = \frac{\mu_o I}{2\pi  r}

This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction

for the case r <r_exterior

in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero

           B₂ = 0

Now we can find the field created by each part

0 < r < r_exterior

          B_total = B₁

          B_total = \frac{\mu_o I}{2\pi  r}  

r > r_exterior

          B_total = B₁ -B₂

          B_total = 0

6 0
3 years ago
You wish to place a spacecraft in a circular orbit around the earth so that its orbital speed will be 4.00×103m/s. What is this
Levart [38]

Answer:

The radius of orbit=2.49\times 10^7 m

Explanation:

We are given that

Orbital speed=4.00\times 10^3m/s

We have to find the radius of orbit of spacecraft.

We know that

Gravitational constant=6.67\times 10^{-11}m^3/kgs^2

Mass of earth=5.972\times 10^{24} kg

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Where G= Gravitational constant

M=Mass of earth

r=Radius of orbit

Substitute the values in the formula

4\times 10^3=\sqrt{\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{r}

Squaring on both sides

16\times 10^6=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{r}

r=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{16\times 10^6}

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7 0
4 years ago
What happens to the volume the gas occupies when the pressure on a gas increases?
taurus [48]
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elena-s [515]

Answer:

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Hope my answer helps you.    Have a nive day!

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Answer:

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