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3 years ago

Why might a balloon, that is inflated almost to its capacity, pop or explode on an extremely warm day?

2 answers:

7 0

On an extremely warm day, the balloon might pop because gases expand the hotter they get, and due to its temperature it is likely to pop if it is, indeed, nearly, if not completely, filled to its capacity. I hope this helps, have a nice day!

nikitadnepr [17]3 years ago

4 0

Explanation:

On an extremely hot day, there will be thermal expansion of the object occur.

With the increase in temperature, the volume of the object will also increase.

In the given problem, balloon is inflated almost to its capacity. But on an extremely hot day, it will expand beyond to its capacity due to the thermal expansion.

Therefore, it might be exploded on an extremely warm day.

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A suspended object A is attracted to a charged object B, can one conclude that A is charged? Explain

irinina [24]

Explanation:

In my view, when the Object A is attracted to a Charged object B. Object B should be Negatively or Positively charged. So Object B should be the Opposite charged according to the Object B

Example =

If Object B is Negatively Charged, the Object A should be Positively Charged

If the Object B is Positively Charged, the Object A should be Negatively Charged

Sometimes it can Mix as a Neutral as well

Hope this Helps

5 0

3 years ago

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

laiz [17]

Answer:

$L=11.3\ kg-m^2/s$

Explanation:

Given that,

Angular speed of a skater, $\omega=3\ rot/s=18.84\ rad/s$

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

$L=I\omega$

Substitute all the values,

$L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s$

So, its angular momentum is equal to $11.3\ kg-m^2/s$ .

8 0

3 years ago

Review the Four Social Errors and Biases in the Highlights area on page 127 in Ch. 4 of THiNK: Critical Thinking and Logic Skill

AysviL [449]

Answer: Provided in the explanation

Explanation:

I have understood that I have been influenced by the 'affinity bias' for quite a while. It caused me to feel fascination and feel better and right about individuals who had comparable intrigue and thought designs. I sort of began to feel this is the one for me based on those likenesses yet later used to be miserable when I comprehended that those similitudes are not many and insufficient consistently. I have begun to beat this inclination by rehearsing self reflection. I have begun to think about what causes me to feel pulled in to somebody and on the off chance that I introspect that it's exclusively founded on likenesses, at that point I cause myself to get that and monitoring that helps in escaping the inclination.

I make an effort not to get into the snare of paradoxes in my own announcements yet I have been forced to bear deceptions during contentions. One deception which I have encountered most is the 'Foul play' error as there have been a great deal of occurrences when individuals began to tear down me and my family when they couldn't win on a contention regarding balanced and rationale.

Basic reasoning causes us in understanding our defects and those issues of our own which keeps us from taking better choices and furthermore forestalls us tackling issues in more successful manners. Through basic reasoning, one can reflect and recognize utilization of deceptions in the contentions and that will help him in deciphering the circumstance from an alternate perspective. He will assess the circumstance better and through appropriate induction, he will have the option to get over those issues in thinking and subsequently, have the option to take better and more powerful choices for taking care of an issue.

The most fascinating idea which I have learned is that of the utilization of reflection or explicitly self reflection to comprehend ourselves better and furthermore to display basic reasoning appropriately. Being said that, I am a still somewhat confused about the manners by which we can reflect appropriately at each circumstance in a successful manner. This disarray remains in light of the fact that most deceptions and predispositions are oblivious in nature while the basic considering aptitudes reflection is cognizant and I meander whether a cognizant ability will have the option to break down and distinguish each oblivious inclinations or whether a few guards will keep a few inclinations covered up.

4 0

3 years ago

A string of 26 identical Christmas tree lights are connected in series to a 120 V source. The string dissipates 73 W. What is th

spin [16.1K]

To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

$R_{eq}= \frac{V}{I}$

Here,

V = Voltage

I = Current

While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

$P=VI \rightarrow I = \frac{P}{V}$

$I =0.608 A$

Applying Ohm's law

$R_{eq} = \frac{120V}{0.608A}$

$R_{eq} = 197.4\Omega$

Therefore the equivalent resistance of the light string is $197.4\Omega$

6 0

3 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago