Answer:
The speed stays constant after the force stops pushing.
Explanation:
Speed always stays constant when the force stops pushing it.
<span>553 ohms
The Capacitive reactance of a capacitor is dependent upon the frequency. The lower the frequency, the higher the reactance, the higher the frequency, the lower the reactance. The equation is
Xc = 1/(2*pi*f*C)
where
Xc = Reactance in ohms
pi = 3.1415926535.....
f = frequency in hertz.
C = capacitance in farads.
I'm assuming that the voltage and resistor mentioned in the question are for later parts that are not mentioned in this question. Reason is that they have no effect on the reactance, but would have an effect if a question about current draw is made in a later part. With that said, let's calculate the reactance.
The 120 rad/s frequency is better known as 60 Hz.
Substitute known values into the formula.
Xc = 1/(2*pi* 60 * 0.00000480)
Xc = 1/0.001809557
Xc = 552.6213302
Rounding to 3 significant figures gives 553 ohms.</span>
Answer:
want
Explanation:
people always want the new thing especially if its technology
Answer:
a)5.88J
b)-5.88J
c)0.78m
d)0.24m
Explanation:
a) W by the block on spring is given by
W= 
kx² = (530)(0.149)² = 5.88 J
b) Workdone by the spring = - Workdone by the block = -5.88J
c) Taking x = 0 at the contact point we have U top = U bottom
So, mg
= kx² - mgx
And, = ( kx² - mgx
)/(mg) = 
]/(0.645x9.8)
= 0.78m
d) Now, if the initial initial height of block is 3
= 3 x 0.78 = 2.34m
then, kx² - mgx - mg =0
(530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0
265x² - 6.321x - 14.8 = 0
a=265
b=-6.321
c=-14.8
By using quadratic eq. formula, we'll have the roots
x= 0.24 or x=-0.225
Considering only positive root:
x= 0.24m (maximum compression of the spring)
