What is the last step in creating an argumentative essay?

A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly ch

IceJOKER [234]

Answer:

Explanation:

Given that,

First Capacitor is 10 µF

C_1 = 10 µF

Potential difference is

V_1 = 10 V.

The charge on the plate is

q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC

q_1 = 100 µC

A second capacitor is 5 µF

C_2 = 5 µF

Potential difference is

V_2 = 5V.

Then, the charge on the capacitor 2 is.

q_2 = C_2 × V_2

q_2 = 5µF × 5 = 25 µC

Then, the average capacitance is

q = (q_1 + q_2) / 2

q = (25 + 100) / 2

q = 62.5µC

B. The two capacitor are connected together, then the equivalent capacitance is

Ceq = C_1 + C_2.

Ceq = 10 µF + 5 µF.

Ceq = 15 µF.

The average voltage is

V = (V_1 + V_2) / 2

V = (10 + 5)/2

V = 15 / 2 = 7.5V

Energy dissipated is

U = ½Ceq•V²

U = ½ × 15 × 10^-6 × 7.5²

U = 4.22 × 10^-4 J

U = 422 × 10^-6

U = 422 µJ

6 0

4 years ago

Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively

Ronch [10]

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

For car 1, we have

$d_1 = u_1 t + \frac{1}{2}a_1 t^2$

where

$u_1 = 7 m/s$ is the initial velocity of car 1

$a_1 = 0.4 m/s^2$ is the acceleration of car 1

So the equation can be rewritten as

$d_1 = 7t + 0.2t^2$

For car 2, we have

$d_2 = u_2 t + \frac{1}{2}a_2 t^2$

where

$u_2 = 4 m/s$ is the initial velocity of car 2

$a_2 = 0.5 m/s^2$ is the acceleration of car 2

So the equation can be rewritten as

$d_2= 5t + 0.25t^2$

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

$d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2$

And solving for t, we find

$2t - 0.05t^2= 0\\t(2-0.05t)=0$

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

$d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

4 years ago

To get a sofa moving which type of friction do you need to overcome?

umka21 [38]

Sliding.
Hope this helps!

5 0

3 years ago

Read 2 more answers

A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is f

AleksAgata [21]

Answer:

Part a)

When rotated about the mid point

$\tau = 0.021Nm$

Part b)

When rotated about its one end

$\tau = 0.042 Nm$

Explanation:

As we know that the angular acceleration of the rod is rate of change in angular speed

so we will have

$\alpha = \frac{\Delta \omega}{\Delta t}$

$\alpha = \frac{2.4 - 0}{3.6}$

$\alpha = 0.67 rad/s^2$

Part a)

When rotated about the mid point

$I = 2mr^2$

$I = 2(0.48)(0.18)^2$

$I = 0.0311 kg m^2$

now torque is given as

$\tau = 0.0311 (0.67)$

$\tau = 0.021Nm$

Part b)

When rotated about its one end

$I = m(2r)^2$

$I = (0.48)(0.36)^2$

$I = 0.0622 kg m^2$

now torque is given as

$\tau = (0.0622)(0.67)$

$\tau = 0.042 Nm$

3 0

3 years ago

Which statement is not true about mass movements?

ololo11 [35]

B i think that the answer so

8 0

3 years ago

Read 2 more answers