Answer:
The velocity
Explanation:
For a mass-spring system, the total mechanical energy is constant during the motion. The total mechanical energy is sum of the elastic potential energy, U, and the kinetic energy, K:
![E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2](https://tex.z-dn.net/?f=E%3DU%2BK%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2%2B%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where
k is the spring constant
x is the displacement
m is the mass
v is the velocity
Since E must remain constant, we see that when x increases, v decreases, and vice-versa. Therefore, when x (the displacement) is at maximum, v (the velocity) is at minimum (more precisely, it is zero).
The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.
To find the answer, we have to know more about the ideal diode.
<h3>
What is an ideal diode?</h3>
- A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
- A semiconductor diode is the kind of diode that is used the most commonly.
- It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.
<h3>How to find the current in ideal diode?</h3>
- Here we have given with the values,
![V_2=65V\\V_1=0V\\R_1=490Ohm.](https://tex.z-dn.net/?f=V_2%3D65V%5C%5CV_1%3D0V%5C%5CR_1%3D490Ohm.)
- We have the expression for current in mA of the ideal diode with forward biased voltage drop as,
![I=\frac{V_2-V_1}{R_1} =\frac{65}{490} =132.6mA](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BV_2-V_1%7D%7BR_1%7D%20%3D%5Cfrac%7B65%7D%7B490%7D%20%3D132.6mA)
Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.
Learn more about the ideal diode here:
brainly.com/question/14988926
#SPJ4
Answer:
<em>The mass of the block is 0.025 kg</em>
Explanation:
<u>Mass-Spring Harmonic Motion</u>
When a mass m is attached to a spring of constant k, they produce a simple harmonic motion which angular frequency is
![\displaystyle w=\sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20w%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
We also know
![w=2\pi f](https://tex.z-dn.net/?f=w%3D2%5Cpi%20f)
which means
![\displaystyle \sqrt{\frac{k}{m}}=2\pi f](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D%3D2%5Cpi%20f)
Squaring
![\displaystyle \frac{k}{m}=4\pi^2 f^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bk%7D%7Bm%7D%3D4%5Cpi%5E2%20f%5E2)
Solving for m
![\displaystyle m=\frac{k}{4\pi^2 f^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m%3D%5Cfrac%7Bk%7D%7B4%5Cpi%5E2%20f%5E2%7D)
We have
![k=10 N/m, f=10/\pi Hz](https://tex.z-dn.net/?f=k%3D10%20N%2Fm%2C%20f%3D10%2F%5Cpi%20Hz)
![\displaystyle m=\frac{10}{4\pi^2 \left (\frac{10}{\pi}\right )^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m%3D%5Cfrac%7B10%7D%7B4%5Cpi%5E2%20%5Cleft%20%28%5Cfrac%7B10%7D%7B%5Cpi%7D%5Cright%20%29%5E2%7D)
Operating
![\displaystyle m=\frac{10}{4\pi^2 \frac{100}{\pi^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m%3D%5Cfrac%7B10%7D%7B4%5Cpi%5E2%20%5Cfrac%7B100%7D%7B%5Cpi%5E2%7D%7D)
Simplifying and computing
![\displaystyle m=\frac{1}{40}=0.025\ kg](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m%3D%5Cfrac%7B1%7D%7B40%7D%3D0.025%5C%20kg)
The mass of the block is 0.025 kg
Answer:
+b±√b² - 4ac /2a
0.6t ± √36-36/2a
Explanation:
Work done = 1/2 mv² where v = (1.2)²
Therefore, 1/2m(1.2)ω mgh
1/2m (1.2)² = 0.4 × m ×10 5
s = 1.44 / 2.4 = 1.44 / 8
S = ut - 1/2gt²
Where u = 1.2
g = 0.9 × 10
Therefore,
1.8 = 1.2v-2t²
2t²c-1.2t+1.8 = 0
t² - 0.6t + 0.9 = 0
0.6t ± √36-36/2a
Solving this further, we make use of the formula
+b±√b² - 4ac /2a