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miskamm [114]
2 years ago
8

if you place 0°c ice into 0°c water in an insulated container, what will happen? Will some ice melt, will more water freeze, or

will neither take place?
Physics
1 answer:
antoniya [11.8K]2 years ago
6 0

Answer:

neither will happen

Explanation:

cause the water is already defreezed

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The electric force between two charges A. increases with distance between the charges B. increases if either one of charges gets
beks73 [17]

Answer:

Option (A) , (b) and (d) are correct option

Explanation:

According to Coulomb's law electric force between two charges is given by

F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}

From the relation we can say that force is directly proportional to magnitude of charges and inversely proportional to distance between them '

So if we increase the distance then force will decrease

Increase if any of the charge get larger

If force is attractive then both the charge will be of different sign and is force is repulsive then both the charges of same sign

From above conclusion we can say that (a), (b) and (d) are correct option

6 0
3 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
3 years ago
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