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kicyunya [14]
3 years ago
15

Neo and Morpheus's masses have gained a velocity (not equal to zero) which means their momentum is now _____ .

Physics
2 answers:
Arte-miy333 [17]3 years ago
6 0
<span>Neo and Morpheus's masses have gained a velocity (not equal to zero) which means their momentum is now based on gravity and friction alone.</span>
laiz [17]3 years ago
3 0

The correct answer to the question is : Their momentum is also increased.

EXPLANATION :

As per the question, the masses of Neo and Morpheus have gained a velocity.

The momentum of a body is defined as the quantity of motion contained in a body. Quantitatively it is the product of mass with velocity.

Mathematically momentum P = m × v

Here, m is the mass of the body, and v is the velocity of the corresponding body.

Hence, momentum of a body increases with the increase in velocity.

As velocity of Neo and Morpheus have increased, so their momentum is also increased.

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What would changing the frequency of a wave do to the wave?
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Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo
LenaWriter [7]

Answer:

6.33\times 10^8\ kg\cdot m/s

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = 33\times \frac{5}{18}=9.17\ m/s

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

                                                             = (6.9\times 10^7\ kg)(9.17\ m/s)

                                                             = 6.33\times 10^8\ kg\cdot m/s

Therefore, the magnitude of ship's momentum is 6.33\times 10^8\ kg\cdot m/s.

6 0
4 years ago
Imagine a universe in which, like in ours, there are two kinds of charges (positive and negative), with the like charges repelli
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Answer:

the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude

Explanation:

In this hypothetical system the electric force is of type

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in this case the force decays to zero much faster,

if we call Fo the force of Coulomb's law

         F₀ = k \frac{q_1 q_2 }{r^2}

assuming the constant k is the same

the relationship between the two forces is

        F / F₀ = 1 / r

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when analyzing this expression the force decays much faster to zero.

In an electric conductor, charges of the same sign may not feel any repulsive force from other charges that are at a medium distance, so there is a probability that some charges are distributed in the volume of the material, this does not happen with coulomb's law

Consequently, the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude

5 0
3 years ago
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iogann1982 [59]

Answer:

Explanation:

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\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2

Solve the system

6 0
3 years ago
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