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lozanna [386]
2 years ago
7

How can i stop loveing you if yo keep saying the things i want to hear

Physics
2 answers:
AleksandrR [38]2 years ago
7 0

Answer:

What do you mean.

Explanation:

xxTIMURxx [149]2 years ago
5 0

Answer:

....

Explanation:

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Will Give Brainlist <br> Check pic for question
Harrizon [31]
Ohms law = v= Ir

V= 0.02 x 4000 = 80v
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3 years ago
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A diver dives off of a raft - what happens to the diver? the raft? how does this relate to newton's third law? action force: ___
kondaur [170]
<span>Actually newtons third law says for every action there is an equal and opposite reaction, Hence here in this case, the diver diving of a raft is the action, after which surely reaction should come in the form where the raft and the driver will rebound with same speed back, and hence here the action force is diving and reaction force is rebounding from the diving place, with same intensity.</span>
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3 years ago
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Work occurs when
andrew-mc [135]
The answer is the FIRST OPTION 
Work occurs when a force is applied to an object and the object moves in the direction of the force applied <span />
6 0
3 years ago
The burning of a log releases the logs chemical_energy into other forms of energy
saul85 [17]

Answer:

When we burn wood we are releasing solar energy, in the form of heat, that has been stored in the wood as chemical energy. The process of photosynthesis converted solar energy, water and carbon dioxide into oxygen and the organic molecules that form the wood, half the weight of which is carbon.

Explanation:

7 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

          $= 2 \times \sqrt{\frac{100}{4}}$

          = 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

    m' = 2m

    Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0
3 years ago
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