v₀ = initial speed as tarzan grabs the vine = 5.3 m/s
v = final speed as the tarzan reach the maximum height = 0 m/s
h = maximum height gained by the tarzan
m = mass of tarzan
using conservation of energy
initial kinetic energy = final kinetic energy + potential energy
(0.5) m v²₀ = (0.5) m v² + m g h
(0.5) v²₀ = (0.5) v² + g h
(0.5) (5.3)² = (0.5) (0)² + (9.8) h
h = 1.43 m
The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
brainly.com/question/14883923
Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.
Hello!
We can use the following equation for calculating power dissipated by a resistor:

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)
Plug in the known values and solve.

The total circuit current at the resonant frequency is 0.61 amps
What is a LC Circuit?
- A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
- These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.
Q =15 = (wL)/R
wL = 30 ohms = Xl
R = 2 ohms
Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance
| Zs | = 30.07 <86.2° ohms
Xc = 1/(wC) = 30 ohms
The impedance of the LC circuit is found from:
Zp = (Zs)(-jXc)/( Zs -jXc)
Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°
I capacitor = 277/-j30 = j9.23 amps
I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps
I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps
Hence, the total circuit current at the resonant frequency is 0.61 amps
To learn more about LC Circuit from the given link
brainly.com/question/29383434
#SPJ4