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Harlamova29_29 [7]
3 years ago
7

In a video game a flying coconut moves at a cost of the last few 20 m/s a positive direction the coconut he stops going to move

in the opposite direction with the cost of the last we have 10 m/s what is the change in velocity of the coconut
Physics
1 answer:
Liula [17]3 years ago
6 0

Answer:

30 m/s

Explanation:

30 m/s I think.

I am too lazy to explain

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Solve the science problem
scZoUnD [109]

Answer:

What is the problem I cant help unless you have the problem.

Explanation:

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The position vector of a particle of mass 1.65 kg as a function of time is given by = (6.00 î + 4.15 t ĵ), where is in meters an
SashulF [63]

Answer:

 L = 41.09 Kg m2 / s      The angular momentum does not depend on the time

Explanation:

The definition of angular momentum is

        L = r x p

Where blacks indicate vectors

Let's apply this definition our case. Linear momentum

      p = m v

Let's replace

      L = m r x v

The given function is

      x = 6.00 i ^ + 4.15 t j ^

We look for speed

     v = dx / dt

     v = 0 + 4.15 j ^

To evaluate the angular momentum one of the best ways is to use determinants

     L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]

      L = m 6 4.15 k ^

The other products give zero

Let's calculate

      L = 1.65 6 4.15 k ^

      L = 41.09 Kg m2 / s

The angular momentum does not depend on the time

7 0
3 years ago
The North Star is located in the Big Dipper. True or False
harina [27]

Answer:

False

Explanation:

It is located in the little dipper whose stars are more faint.

4 0
3 years ago
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How many hydrogen atoms are there in the following compound: 4CHA (2 points)
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Answer:

answer a, 4

Explanation:

when the 4 is before the compound it applies to the whole compound

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A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

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