Answer:
circuit sketched in first attached image.
Second attached image is for calculating the equivalent output resistance
Explanation:
For calculating the output voltage with regarding the first image.
![Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V](https://tex.z-dn.net/?f=Vout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5B%2F%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DVout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5C%5CVout%20%3D%205%20%5Cfrac%7B2%7D%7B5%7D%20%3D%202%20V)
For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.
so.
Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.
if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:
so.
Question
Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.
Answer:
Explanation:
The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,
Here, l is the length of the baton.
Substitute 0.120 kg for m, 3 rads/s for ![\omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}](https://tex.z-dn.net/?f=%5Comega%5B%5Ctex%5D%20and%200.8%20m%20for%20l%20%5Btex%5D%5Cbegin%7Barray%7D%7Bc%7D%5C%5CL%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7Dm%7Bl%5E2%7D%5Comega%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7D%5Cleft%28%20%7B0.120%7B%5Crm%7B%20kg%7D%7D%7D%20%5Cright%29%7B%5Cleft%28%20%7B%7B%5Crm%7B80%7D%7D%7B%5Crm%7B.0%20cm%7D%7D%7D%20%5Cright%29%5E2%7D%7B%5Cleft%28%20%7B%5Cfrac%7B%7B1%20%5Ctimes%20%7B%7B10%7D%5E%7B%20-%202%7D%7D%7B%5Crm%7Bm%7D%7D%7D%7D%7B%7B1%7B%5Crm%7B%20cm%7D%7D%7D%7D%7D%20%5Cright%29%5E2%7D%5Cleft%28%20%7B%7B%5Crm%7B3%7D%7D%7B%5Crm%7B.00%20rad%2Fs%7D%7D%7D%20%5Cright%29%5C%5C%5C%5C%20%3D%200.0192%7B%5Crm%7B%20kg%7D%7D%20%5Ccdot%20%7B%7B%5Crm%7Bm%7D%7D%5E%7B%5Crm%7B2%7D%7D%7D%7B%5Crm%7B%2Fs%7D%7D%5C%5C%5Cend%7Barray%7D)
The angle of refraction would be further less
The velocity of the aircraft relative to the ground is 240 km/h North
Explanation:
We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.
Mathematically:
where
v' is the velocity of the aircraft relative to the ground
v is the velocity of the aircraft relative to the air
is the velocity of the air relative to the ground.
Taking north as positive direction, we have:
v = +320 km/h
(since the air is moving from North)
Therefore, we find
(north)
Learn more about vector addition:
brainly.com/question/4945130
brainly.com/question/5892298
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<span>Answer:The weight of the door creates a CCW torque given by
Tccw = 145 N*3.13 m / 2
You need a CW torque that's equal to that
Tcw = F*2.5 m*sin20</span>
