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Naddik [55]
3 years ago
6

Which question would most likely fill in the blank

Physics
1 answer:
Citrus2011 [14]3 years ago
8 0
Definitely not the last 2. My bet is on the first option. If it is wrong don't hit me please...
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The ray diagram shows a vase that is placed beyond the center of curvature of a concave mirror.
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The image will form in the vicinity of F. Its nature will be small and inverted 
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3 years ago
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The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for
abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

5 0
3 years ago
What is the correct equation for calculating the average atomic mass for 3 isotopes? (pls be 100%of your answer pls no guessing)
mariarad [96]

<u>Answer:</u>

<em>The correct equation for measuring the average microscopic weight  for 3 isotopes is multiply the rate of abundance by each weight and add them.</em>

<u>Explanation:</u>

To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.

  • Take the correct weight of each isotope (that will be in decimal form)
  • Multiply the weight of each isotope by its abundance
  • Add each of the results together.

<em>This gives the required  average microscopic weight of the three isotopes.</em>

3 0
3 years ago
A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte
musickatia [10]

Answer:

Number of electrons, n=5.62\times 10^{21}

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, I=\dfrac{q}{t}

I\times t=n\times e

n=\dfrac{It}{e}

e is the charge of an electron

n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}

n=5.62\times 10^{21}

So, the number of electrons pass through the resistor is 5.62\times 10^{21}. Hence, this is the required solution.

6 0
3 years ago
Which equation could not be used to determine straight line acceration?
uranmaximum [27]

Answer:

the answer for the question is the last option

5 0
3 years ago
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