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3 years ago

14

The lens system within a camera forms a _______ on the sensor. A. virtual, upright image B. real, inverted image C. r

eal, upright image D. virtual, inverted image

2 answers:

Rashid [163]3 years ago

7 0

B. The image on the film or sensor is real and inverted.

Lisa [10]3 years ago

5 0

The answer is virtual and inverted

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7nadin3 [17]

Answer:

2.5 m/s²

Explanation:

Using the formula, v = u + at ( v = Final velocity; u = Initial velocity; t = Time; a = Acceleration)

25 = 0 + 10a

a = 25/10 = 2.5 m/s²

8 0

3 years ago

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:

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

<h3>a)</h3>

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

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:

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

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

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

<h3>b)</h3>

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

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

what is the energy of a single photon of ultraviolet light with a wavelength of 30.0 nm? 1 nm=10^-9 m

Arada [10]

From the planks equation
E=hv
V= c/ wave length
V= 3×10^8/30×10^-9
=1×10^16
E= hv
6.63×10^-34×1×10^16
= 6.63×10^-18

4 0

3 years ago

According to Lenz's law, the induced current in a circuit always flows to oppose the external magnetic field through the circuit

Makovka662 [10]

Answer:

True.

Explanation:

According to Lenz's law, the induced current in a circuit always flows to oppose the external magnetic field through the circuit. This statement is true.

The Faraday's law of induction is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

Here, negative sign shows that the direction of induced emf is such that opposes the changing current that is its cause.

Hence, the statement is true.

3 0

3 years ago

What is an electrical charge how does it create electrical current

AveGali [126]

Answer:

Explanation:

The moving charged particles in an electric current are called charge carriers. In metals, one or more electrons from each atom are loosely bound to the atom, and can move freely about within the metal. These conduction electrons are the charge carriers in metal conductors.

The flow of electrons in a direction is known as electric current. The tendency of attraction between the positive and negative charges makes electric current flow through a wire

7 0

3 years ago