Please help me answer this it is due by Sunday night.

A 2,000 kg car is parked at the top of a 30 m high hill. what is its potential energy?

Phoenix [80]

The PE for this question will be 588,000 because we take the mass (2,000 kg), multiply it by 9.8 which is Gravitational Acceleration and then multiply that by the height (30 meters)

6 0

3 years ago

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

If we warm a volume of air, it expands. Does it then follow that if we expand a volume of air, it warms? Explain.

Ivahew [28]

Answer:

nope don't think so

Explanation:

the heat causes the molecules to move faster therefore expanding in watever it the air is in

3 0

3 years ago

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How do you know the speed of an electromagnetic wave in a vacuum?

ycow [4]

Electromagnetic waves need no matter to travel - they can travel through empty space (a vacuum). In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - which is the fastest speed possible. ...
Light traveling value through an optical Fibre is, 2 x 108 m/s. Hope that helps.

5 0

3 years ago

If the magnitude of the electric field in air exceeds roughly 3 âœ• 106 n/c, the air breaks down and a spark forms. for a two-di

Vlad1618 [11]

Answer: 39.8 μC

Explanation:

The magnitude of the electric field generated by a capacitor is given by:

$E = \frac{V}{d}$

d is the distance between the plates.

For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.

$C =\frac{\epsilon_o A }{d}$

where A is the area of the plate and ε₀ is the absolute permittivity.

substituting, we get

$E = \frac{Q}{\epsilon_o A}$

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.

radius of the plates of the capacitor, r = 69 cm = 0.69 m

Area of the plates, A = πr² = 1.5 m²

Thus, the maximum charge that can be placed on disks without a spark is:

Q = E×ε₀×A

⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.

8 0

3 years ago