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FrozenT [24]
2 years ago
13

How do the positions of the sun and moon affect what people do?

Physics
1 answer:
Verdich [7]2 years ago
6 0

-- The position of the sun was originally the primary influence in determining
when people went to sleep and when they woke up.  Although it no longer
directly influences us, that pattern is so deeply ingrained in our make-up
that our behavior still largely coincides with the positions of the sun.

-- The position of the Moon was originally the primary influence in determining
the cycle of human female physiology.  Although it no longer directly influences
us, that pattern is so deeply ingrained in human make-up that the female cycle
still largely coincides with the positions of the Moon.



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Write the equation that links current, potential difference, and resistance
mrs_skeptik [129]

You can write the equation in 3 different ways, depending on which quantity you want to be the dependent variable.  Any one of the three forms can be derived from either of the other two with a simple algebra operation.  They're all the same relationship, described by "Ohm's Law".

==> Current = (potential difference) / (resistance)

==> Potential difference = (current) x (resistance)

==> Resistance = (potential difference) / (resistance)

4 0
3 years ago
A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the
SOVA2 [1]

Answer:

Johnny created an electromagnet out of a solenoid (a coil of wires with 20 loops), an iron core (of 1 nail), and a single 9 V battery. When Johnny does this, he creates a small magnetic field that allows him to pick up 2 paper clips. Using a CER format, explain to Johnny three things he could change that would increase the strength of his magnetic field and why each change increases the magnetic field. You may want to write three paragraphs to make this easier for the reader to understand

5 0
2 years ago
A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear
JulijaS [17]
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
5 0
3 years ago
Calculate the acceleration of a 270000-kg jumbo jet just before takeoff when the thrust on the aircraft is 160000 N .
Radda [10]

Answer:

<h3>The answer is 0.59 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{160000}{270000}  =  \frac{16}{27}  \\  = 0.592592...

We have the final answer as

<h3>0.59 m/s²</h3>

Hope this helps you

7 0
3 years ago
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