Ask question

Ask question

All categories

3 years ago

7

Problem 1: Given: 14 Coulombs of charge flow in an electrical circuit. The charge is pushed by a voltage difference if 8 volts.

A) Calculate the amount of work done in volt. Coulombs. B.) Calculate the work done in Joules.

Advanced Placement (AP)

1 answer:

Sedbober [7]3 years ago

4 0

B.) Calculate the work done in Joules.

You might be interested in

HEYY I NEED HELP WITH AN AP CALC ASSIGNMENT ASAP

Vladimir79 [104]

CALCULATOR PART

1. The area of R + S is unsigned, meaning you want to find

$\displaystyle\int_a^b\left|f(x)-g(x)\right|\,\mathrm dx$

where $[a,b]$ is the interval between the leftmost and rightmost intersections of $f$ and $g$ .

First use your calculator to find these intersections:

$\cos x=\dfrac{x+1}3\implies x\approx-3.64,x\approx-1.86,x\approx0.889$

so that $a=-3.64$ and $b=0.889$ . Now compute the integral using your calculator:

$\displaystyle\int_a^b\left|f(x)-g(x)|\,\mathrm dx\approx1.662$

2. The volume, using the washer method, is given by the integral

$\displaystyle\pi\int_{-1.86}^{0.889}(|2-g(x)|^2-|2-f(x)|^2)\,\mathrm dx\approx12.078$

3. A circle of radius $r$ has area $\pi r^2$ ; a semicircle with the same radius thus has area $\frac{\pi r^2}2$ . Each cross section of this solid is a semicircle whose diameter is the vertical distance between $f(x)$ and $g(x)$ , or $|f(x)-g(x)|$ . In terms of the diameter $d=2r$ , the area of each semicircle would be $\frac{\pi d^2}8$ . Then the volume of the solid is

$\displaystyle\frac\pi8\int_{-3.64}^{-1.86}|f(x)-g(x)|^2\,\mathrm dx\approx0.0425$

NON-CALCULATOR PART

4. The mean value theorem says that for a function $F$ continuous on an interval $[a,b]$ and differentiable on $(a,b)$ , there is some $c\in(a,b)$ such that

$F'(c)=\dfrac{F(b)-F(a)}{b-a}$

If this $F$ happens to be an antiderivative of $f$ , then we end up with

$f(c)=\displaystyle\frac1{b-a}\int_a^bf(x)\,\mathrm dx$

$\cos x$ is continuous and differentiable everywhere, so the MVT applies. We have $F'(x)=f(x)=\cos x$ , so the MVT tells us there is some $c\in[0,\pi$ such that

$\cos c=\dfrac{\sin\pi-\sin0}{\pi-0}=0$

That is, the average value of $f(x)$ on $[0,\pi]$ is 0. The MVT says there is some $c$ in the interval such that the function takes on the average value itself; this happens for $c=\frac\pi2$ .

5. This question seems to be incomplete...

5 0

4 years ago

Clients expect that information shared during a session will

Pani-rosa [81]

Clients expect that information shared during a session will be confidential.

Depending on their problem, many people don't like to share what's going on with them except someone who can help them.

Hope this helps you understand!!
Please make brainliest to help me!!
Thanks!!

5 0

4 years ago

Read 2 more answers

What is the difference between an ecosystem and biome

Gre4nikov [31]

Ecosystem includes all of the biotic and abiotic factors that are found in a given environment. Biome is a collection of different ecosystems which share similar climate conditions.

7 0

3 years ago

Hello, I need help with a calculus FRQ. My teacher has given a hint that this last part has to do with the intermediate value th

lesya [120]

Answer:

Yes, at a time t such that (√2)/2 ≤ t ≤ 2.

Explanation:

To answer the question

Therefore, where the domain of the function is the set of all real numbers x for which f(x) is a real number we have

For Chloe's velocity

$C(t) = t\times e^{4-t^2} \ for \ 0\leq t\leq 2$

Finding the boundaries of the function gives;

$0\times e^{4-0^2} = 0$ and $2\times e^{4-2^2} = 2$

At t = 1, we have $1\times e^{4-1^2} = e^{3} = 20.086$

We find the maximum point as follows;

$\frac{\mathrm{d} \left (t\times e^{4-t^2} \right )}{\mathrm{d} x}=0$

From which we have;

$\frac{\mathrm{} e^{4-t^2} - t\times e^{4-t^2} \times2\times t }{(e^{4-t^2} )^2}=0$

$e^{4-t^2} - t\times e^{4-t^2} \times2\times t }=0$

$e^{4-t^2}(1 - t\times2\times t })=0\\e^{4-t^2}(1 - 2\times t^2 })=0\\$

$e^{4-t^2}=0$ or $(1 - 2\times t^2 })=0$

∴ 1 = 2·t² and from which t = (√2)/2

Hence the function C(x) is decreasing from t = (√2)/2 to t = 2

For Brandon

For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5

1 ≤ f(x) ≤ 1.5

Given that the function B(t) is differentiable, therefore, continuous, there exists a point at which the function C(t) and B(t) intersects given that;

For 0 ≤ t ≤ (√2)/2, 0 ≤ C(t) ≤ 23.416 for (√2)/2 < t ≤ 2, 23.416 > C(t) ≥ 2

and for 0 ≤ t ≤ 0 1 ≤ B(t) ≤ 8 and for 1 < t ≤ 2, 8 > B(t) ≥ 1.5

Therefore, the curves intersect at in between (√2)/2 ≤ t ≤ 2.

8 0

3 years ago

Pre ap geometry I have no idea how to do this please help me

Yuki888 [10]

10. The two sides are equal because that is the definition of a midpoint so
8x+11=12x-1
-8x -8x
11=4x-1
+1 +1
12=4x
12/4=3
X=3

8 0

3 years ago