Well, first of all, the truck's velocity is constantly changing, not 'uniform'.
Velocity consists of speed and direction. So, even if the truck's speed is
constant, its direction keeps changing as long as it's on a circular curve,
so its velocity is constantly changing.
The force needed to keep a mass moving in a circle is
F = (mass) x (speed)² / (radius)
3300 N = (1600 kg) (13 m/s)² / R
3300 kg-m/s² = (1600 kg) (169 m²/s²) / R
R = (1600 kg) · (169 m²/s²) / (3300 kg·m/s²)
= (1600 · 169 / 3300) meters
= 81.9 meters
Answer:
C. 5 miles/ minute north east
Explanation:
magnitude = 10 mi / 2 min = 5 mi/min
Velocity is a vector
A vector always has both a magnitude and direction
The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,
![Xc = \frac{1}{2\pi fC}](https://tex.z-dn.net/?f=Xc%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%20fC%7D)
where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?
![Xc' = \frac{1}{2\pi f'C}](https://tex.z-dn.net/?f=Xc%27%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%20f%27C%7D)
![= \frac{1}{2\pi 2f C}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%202f%20C%7D)
![= \frac{1}{2} (\frac{1}{2\pi fC} )\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B1%7D%7B2%5Cpi%20fC%7D%20%29%5C%5C)
![Xc' = \frac{1}{2} Xc](https://tex.z-dn.net/?f=Xc%27%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20Xc)
Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
Learn more about capacitive reactance here:
brainly.com/question/23427243
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The choice c. 100 cm / s²
__o_o_
![a \: max = {w}^{2} A = {\pi}^{2} \times 10 = 10 \times 10 = 100 \: \frac{cm}{ {s}^{2} }](https://tex.z-dn.net/?f=a%20%5C%3A%20max%20%3D%20%20%7Bw%7D%5E%7B2%7D%20A%20%3D%20%20%7B%5Cpi%7D%5E%7B2%7D%20%20%5Ctimes%2010%20%3D%2010%20%5Ctimes%2010%20%3D%20100%20%5C%3A%20%20%5Cfrac%7Bcm%7D%7B%20%7Bs%7D%5E%7B2%7D%20%7D%20)
I hope I helped you^_^
Answer:
For each second the position increased by 10 m