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Margarita [4]
3 years ago
14

Tell me about your experiences with science​

Physics
2 answers:
skelet666 [1.2K]3 years ago
8 0

Answer:

so good

Explanation:

<h2>perfect nice and explody too much</h2>
atroni [7]3 years ago
7 0

Answer:

Well i once did an expiremnet and it exploaded in my face lol

Explanation:

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Nodes give a vertebrate’s back flexibility.<br> A. True<br> B. False
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No B. its false<span> they do not. Lyphnodes are a permant part of the body and you have them in your Neck</span>
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3 years ago
Read 2 more answers
Two small, identical conducting spheres repel each other with a force of 0.045 N when they are 0.15 m apart. After a conducting
Levart [38]

Answer:

q_1 = \pm 1.68 \times 10^{-7} C

q_2 = \pm 6.68 \times 10^{-7} C

Explanation:

As we know that the force between two small spheres is given as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 , q_2 = charges on two small spheres

r = distance between two spheres = 0.15 m

now the force between them is given as

0.045 = \frac{(9\times 10^9)(q_1q_2)}{0.15^2}

q_1q_2 = 1.125 \times 10^{-13}

now when two spheres are connected together then the charge on them is equally divided

q = \frac{q_1+q_2}{2}

now the force between them is given as

F = \frac{k(\frac{q_1+q_2}{2})^2}{0.15^2}

0.070 = \frac{(9\times 10^9)(\frac{q_1+q_2}{2})^2}{0.15^2}

q_1 + q_2 = 8.37\times 10^{-7}

so here we have

q_1 = \pm 1.68 \times 10^{-7} C

q_2 = \pm 6.68 \times 10^{-7} C

5 0
3 years ago
What happens inside someone's body when they danve
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The body releases endorphins
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What is the first and second law of thermodynamics
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3 years ago
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Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A
Crazy boy [7]

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction. </em>

<em> How much work W does the motor do on the platform during this process?  Enter your answer in joules to four significant figures.</em>

The amount of work done by the motor is given by:

W=\Delta K

W= 1/2*I*\omega f^2-1/2*I*\omega o^2

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

\omega f^2=\omega o^2+2*\alpha*\theta

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

\tau=I*\alpha

\alpha=\tau/I     =>     \alpha = 0.5rad/s^2

Now we can calculate the final velocity:

\omega f = 8.68rad/s

Finally, we calculate the total work:

W= 1/2*I*\omega f^2 = 1883.56J

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":

W = 1884J

3 0
3 years ago
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