PLEASE HELP TIME SENSITIVE:)

Calculate the linear acceleration of a car, the 0.220-m radius tires of which have an angular acceleration of 13.0 rad/s2. Assum

Anuta_ua [19.1K]

Answer:

The linear acceleration of the car is 2.86m/s².

Explanation:

First of all, we have that the linear acceleration of the car and the tires is the same, because it moves ahead without disassembling itself. So, we only have to calculate the linear acceleration of the tires.

Next, it is known that the linear and angular acceleration are related by the formula:

$a=\alpha R$

Finally, we plug in the given values in order to calculate the acceleration:

$a=(13.0rad/s^{2})(0.220m)=2.86m/s^{2}$

It means that the linear acceleration of the car is 2.86m/s².

6 0

4 years ago

How does a electron travel through a closed circuit?

Murljashka [212]

The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.

3 0

4 years ago

WILL MARK BRAINLIEST IF THE ANSWER IS CORRECT Vinegar and baking soda react to form sodium acetate (NaC2H3O2), water (H2O), and

TEA [102]

D. 3, hope this helps

4 0

3 years ago

Please help solve , picture attached

Alexxx [7]

14 you have to multiply the 7 by the 2

3 0

3 years ago

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

a)

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago