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Thepotemich [5.8K]

3 years ago

11

An initially uncharged 3.67 μF capacitor and a 8.01 k Ω resistor are connected in series to a 1.50 V battery that has negligible

internal resistance. What is the initial current in the circuit, expressed in milliamperes?

1 answer:

inn [45]3 years ago

8 0

Answer:

The initial current in the circuit is 0.1873 mA

Explanation:

Given;

capacitance is given as 3.67 x 10⁻⁶ F

resistance is given as 8010 Ω

voltage across the circuit is 1.5 V

Since the capacitor is initially uncharged, the capacitive reactance is zero.

From ohms law;

Voltage across the circuit is directly proportional to the opposition to the flow of current.

V = IR

I = V/R

I = 1.5/8010

I = 0.0001873 A = 0.1873 mA

Therefore, the initial current in the circuit is 0.1873 mA

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Students have 4 different pairs of sunglasses, each with lenses where the polarization filter is oriented in different direction

AysviL [449]

Answer:

the lens you must select has an angle of 143º measured with respect to the horizontal, this angle is 53º with respect to the vertical.

Explanation:

The glare is caused by the reflection of light in the water, the polarization of the reflected light is polarized in a direction parallel to the surface of the water, the polarization is total for the angles

n = tan $\theta_{p}$

\theta_{p} = tan⁻¹ n

the refractive index for seawater is 1.33

\theta_{p}= tan⁻¹ 1.33

\theta_{p} = 53º

for this angle the light is totally polarized, for the other angles the polarization is partial.

Based on this, the lenses must eliminate this polarization, so its polarization direction must have 90º with respect to this polarization,

\theta_{lens} = 53 +90

\theta_{lens}= 143º

Therefore, the lens you must select has an angle of 143º measured with respect to the horizontal, this angle is 53º with respect to the vertical.

A lens that could work is one that is polarized 45º with respect to the vertical.

6 0

3 years ago

2.List two changes an unbalanced force can cause.

Evgen [1.6K]

Answer:

By applying an unbalanced force, you can change the motion of an object. Unbalanced forces can make an object at rest start moving, make a moving object stop, or change the direction and speed of the object.

Explanation:

3 0

3 years ago

three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}$

Simplfying and solving for $q_1$

$\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}$

$\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}$

$\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}$

5 0

3 years ago

The acceleration of an object would increase if there was an increase in the

xxMikexx [17]

As per Newton's II law we know that

$F_{net} = ma$

here we know that

$F_{net} = F_{ap} - F_f$

so here we will have

$a = \frac{F_{ap} - F_f}{m}$

so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.

So here correct answer will be

<em>A) force on the object.</em>

4 0

3 years ago

A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha

Kisachek [45]

Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

$T = \frac{1}{f}$

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

$T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s$

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:

$A = \frac{20 cm}{2} = 10 cm$

Therefore, the amplitude is 10 cm.

I hope it helps you!

5 0

3 years ago