2. the energy of the pendulum as it swings up

In an oscillating L C circuit, the maximum charge on the capacitor is 1.5 × 10 − 6 C and the maximum current through the inducto

Andre45 [30]

Answer:

T= 1.71×10^{-3} sec= 1.71 mili sec

t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec

Explanation:

First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)= 5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C

a) I(t) = -ωqosin(ωt+φ)

⇒Io= ωqo

⇒ω= Io/qo

also we know that T= 2π/ω

⇒T= $2\pi\frac{q_0}{I_0}$

now putting the values we get

= $2\pi\frac{1.5\times10^{-6}}{5.5\times10^{-3}}$

= 1.71×10^{-3} sec

b) note that the time $t_{fc}$ it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is

$t_{fc}= \frac{T}{4}$

$t_{fc}= \frac{ 1.71\times10^{-3} }{4}$

t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec

5 0

3 years ago

A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?

Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

$H_m = \frac{u^2sin^2 \theta}{2g}$

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

$H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta = sin^{-1}(0.2123)\\\\\theta = 12.26^0$

Therefore, the angle of projection is 12.26⁰.

6 0

2 years ago

An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t

Lynna [10]

Answer:

2.5 m

Explanation:

Weight of billboard worker = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension in rope = 550 N

The sum of the torques will be

$-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m$