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9 days ago

7

Two devices with capacitances of 25 μf and 5.0 μf are each charged with separate 120 v power supplies. calculate the total energ

y stored in the two capacitors. your answer submit

2 answers:

Ostrovityanka [3K]9 days ago

6 0

Answer:

The cumulative energy contained in both capacitors amounts to 0.0018J

Explanation:

The formula used for determining the total energy between two capacitors is 1/2CV²

Energy stored in the First Capacitor = 1/2 × 25× 10⁻⁶f × (120)²

Energy stored equals 0.0015J

Energy stored in the second Capacitor = 1/2 × 5× 10⁻⁶f × (120)²

Energy stored equals 0.0003J

Total Energy stored equals the energy stored in Capacitor 1 plus the energy in Capacitor 2

= 0.0015J + 0.0003J

= 0.0018J

kicyunya [3.1K]9 days ago

3 0

The energy held in a capacitor can be calculated using
$U= \frac{1}{2} CV^2$
, where C represents capacitance and V denotes the voltage applied across the capacitor.

We will compute the energy for the first capacitor:
$U_1 = \frac{1}{2} (25\cdot 10^{-6}F)(120 V)=1.5 \cdot 10^{-3}J$

Next, let's find the energy for the second capacitor:
$U_2 = \frac{1}{2} (5 \cdot 10^{-6}F)(120 V)=3 \cdot 10^{-4}J$

Thus, the combined energy stored in both capacitors amounts to
$U=U_1 +U_2 = 1.8 \cdot 10^{-3}J$

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A uniform magnetic field makes an angle of 30o with the z axis. If the magnetic flux through a 1.0 m2 portion of the xy plane is

Yuliya22 [3221]

Response:

(b) 10 Wb

Clarification:

Given;

angle of the magnetic field, θ = 30°

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Dee is on a swing in the playground. the chains are 2.5 m long, and the tension in each chain is 450 n when dee is 55 cm above t

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23 days ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [2943]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

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$y=7.39\ m$

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