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Alex Ar [27]
2 years ago
14

[sorry im doing this again] When does an electrically charged object attract another object?

Physics
1 answer:
mojhsa [17]2 years ago
8 0
B Since when each object is negatively charged it becomes a electronic. Basically
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Drag the tiles to the correct boxes to complete the pairs.
marishachu [46]

Answer:

Air pollution---> smog

Water pollution---> eutrophication

Land pollution---> contaminated soil

Light pollution---> sky glow

7 0
3 years ago
An enemy sub is approaching a us submarine at 25.0 km/hr that is waiting in ambush. the us submarine needs to know the exact pos
Illusion [34]
Speed of the Enemy sub .....> <span> 25.0 km/hr which is (25/3.6) m/s = 6.94 m/s

We use the expression Speed = Distance/ time to find the distance

D = Speed*Time = 6.94 m/s * 3.5  = 24.305 m

The enemy sub will be 24.305 m closer 
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3 0
3 years ago
I need help real quick. Please help me!!
dimaraw [331]
<h3><u>Answer;</u></h3>

D) Standing wave

<h3><u>Explanation;</u></h3>
  • Standing wave also called stationary wave  is a wave which oscillates in time but whose peak amplitude profile does not move in space.
  • A standing wave pattern is a vibrational pattern created within a medium when the vibrational frequency of the source causes reflected waves from one end of the medium to interfere with incident waves from the source.
  • Examples of standing waves include the vibration of a violin string and electron orbitals in an atom.
4 0
3 years ago
What type of weather would be in the middle of the picture, where the barometric pressure is 1000 millibars?
UkoKoshka [18]
It would be D. Hope that helped :)
4 0
3 years ago
A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

5 0
2 years ago
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