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professor190 [17]
3 years ago
9

two small balls are suspended on parallel threads of the same length so that they touch each other in the vertical position. the

mass of the first ball is 0.4kg and the mass of the second ball is 200g. the first ball is deflected so that its centre of mass rises to a height of 6cm and it is then released. to what maximal height will the 200g ball rise if the collision is elastic?
Physics
1 answer:
Elden [556K]3 years ago
5 0

Answer: 12cm

Explanation:

Since first ball is raised to a height h1= 6cm, it gains a potential energy(PE1). On release, this PE1 is converted to kinetic Energy (KE1). This KE1 impact it energy on ball 2 ans it rise also. Since collision is elastic.

KE1 = KE2 ....> PE1 = PE2

m1 ×g × h1 = m2 × g× h2

h2 = m1×h1/m2

h2 = 0.4 × 0.06/0.2

h2 = 0.12m = 12cm

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Charge flows from low potential to high potential.<br> O True<br> or<br> O False
inn [45]

Answer:

False

Explanation:

Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).

Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).

Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.

8 0
3 years ago
2. Can you place three forces of 5g, 6g, and 12g so they are in equilibrium. Justify your answer.
Bond [772]

Answer:

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

Explanation:

Equilibrium means their sum must be zero.

Here the forces are 5g, 6g, and 12g.

For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.

Here

        Magnitude of largest force = 12 g

        Sum of magnitudes of other forces = 5g + 6g = 11g

       Magnitude of largest force >   Sum of magnitudes of other forces

So this forces cannot form equilibrium.

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

4 0
3 years ago
Easy quiz easy points 7
Darya [45]
The answer is 2: hush
5 0
3 years ago
Read 2 more answers
The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, w
Viktor [21]

Answer:

68 readings.

Explanation:

We need to take this problem as a statistic problem where the normal distribution table help us.

We can start considerating that X is the temperature of the solution, then

0.9 = P(|\bar{x}-\mu|

0.9 = P(\frac{|\bar{x}-\mu|}{\frac{\sigma}{\sqrt{n}}}

0.9 = P(|Z|

For a confidence level of 90% our Z_{critic} is 1.645

Therefore,

\frac{0.1}{\frac{\sigma}{\sqrt{n}}} = 1.645

Substituting for \sigma = 5 and re-arrange for n, we have that n is equal to

n=(\frac{1.645\sigma}{0.1})^2

n=\frac{(1.645)^2(0.5)^2}{0.1^2}

n=67.65

n=68

We need to make 68 readings for have a probability of 90% and our average is within 0.1\°\frac

3 0
3 years ago
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svp [43]
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5 0
3 years ago
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