Answer:
The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.
Explanation:
Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector
y
so that
;
to get how far we use Pythagorean theorem so
so that ![R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)](https://tex.z-dn.net/?f=R%3D%5Csqrt%7B0.375%5E%7B2%7D%2B5.01%5E%7B2%7D%20%7D%20%3D5.02%20%28Km%29)
The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.
So when you stand on the surface of this particular planet, you feel a force of gravity that is
(1/2) / (3²)
of the force that you feel on the surface of the Earth.
That's <em>(1/18)</em> as much as on Earth.
The acceleration of gravity there would be about <em>0.545 m/s²</em>.
This is about 12% less than the gravity on Pluto.
Answer:
D. If a home were wired in series, every light and appliance would have to be turned on in order for any light or appliance to work.
Explanation:
In a series circuit, all the appliances are connected on the same branch of the circuit, one after the other. This means that the current flowing throught them is the same. However, this means also that if one of the appliance is turned off (so, its switch is open), that appliance breaks the circuit, so the current can no longer flow through the other appliances either.
On the contrary, when the appliances are connected in parallel, they are connected in different branches, so if one of them is switched off, the other branches continue working unaffacted by it.
Answer:
The inductance of the inductor is 35.8 mH
Explanation:
Given that,
Voltage = 120-V
Frequency = 1000 Hz
Capacitor ![C= 2.00\mu F](https://tex.z-dn.net/?f=C%3D%202.00%5Cmu%20F)
Current = 0.680 A
We need to calculate the inductance of the inductor
Using formula of current
![I = \dfrac{V}{Z}](https://tex.z-dn.net/?f=I%20%3D%20%5Cdfrac%7BV%7D%7BZ%7D)
![Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%7BR%5E2%2B%28L%5Comega-%5Cdfrac%7B1%7D%7BC%5Comega%7D%29%5E2%7D)
Put the value of Z into the formula
![I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BV%7D%7B%5Csqrt%7BR%5E2%2B%28L%5Comega-%5Cdfrac%7B1%7D%7BC%5Comega%7D%29%5E2%7D%7D)
Put the value into the formula
![0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}](https://tex.z-dn.net/?f=0.680%3D%5Cdfrac%7B120%7D%7B%5Csqrt%7B%28100%29%5E2%2B%28L%5Ctimes2%5Cpi%5Ctimes1000-%5Cdfrac%7B1%7D%7B2%5Ctimes10%5E%7B-6%7D%5Ctimes2%5Cpi%5Ctimes1000%7D%29%5E2%7D%7D)
![L=35.8\ mH](https://tex.z-dn.net/?f=L%3D35.8%5C%20mH)
Hence, The inductance of the inductor is 35.8 mH
Answer:
The magnitude = 10.30 m
The direction of the vector proceeds at angle of 119.05°
Explanation:
Given that:
A vector
has component
= -5 m and
= 9 m
The magnitude of vector
can be represented as:
= ![\sqrt{A_x^2 + A_y^2}](https://tex.z-dn.net/?f=%5Csqrt%7BA_x%5E2%20%2B%20A_y%5E2%7D)
= ![\sqrt{(-5)^2 + (9)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-5%29%5E2%20%2B%20%289%29%5E2%7D)
= ![\sqrt{25 + 81}](https://tex.z-dn.net/?f=%5Csqrt%7B25%20%2B%2081%7D)
= ![\sqrt{106}](https://tex.z-dn.net/?f=%5Csqrt%7B106%7D)
= 10.30 m
If we make
an angle
with y- axis:
Then; tan
= ![\frac{A_x}{A_y}](https://tex.z-dn.net/?f=%5Cfrac%7BA_x%7D%7BA_y%7D)
tan
= ![\frac{5}{9}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B9%7D)
tan
= 0.555
= tan⁻¹ (0.555)
= 29.05°
Angle with positive x-axis = 90 +
= 90° + 29.05°
= 119.05°