Sophia wanted to estimate the product 73 x 12 so she made an estimate of 70x 10. Would her estimate be greater or near the actual estimate?

Answers

Answer 1

Answer: her estimate would not be greater, so it would be near the actual estimate. the numbers 70 and 10 are smaller than the numbers 73 and 12, so it is not possible for 70 x 10 to be more than 73 x 12 is. i hope this helps!!

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Assume the electric field has a magnitude of 59.0 m V , and the particle has a net positive charge of Q=+5.40 C. Calculate the magnitude of the electric potential difference, \vert \Delta V \vert = \vert V_B - V_A \vert∣ΔV∣=∣V B −V A ∣.

Answers

Answer:

V = 1.69 * 10^6 V

Explanation:

Parameters given:

Electric field, E = 59V/m

Charge, q = 5.40C

We need to first find the distance between the electric charge and the point of consideration to be able to find the Electric potential difference.

Electric field is given as:

E = (kq/r^2)

k = Coulombs constant

=> r^2 = kq/E

=> r^2 = (9 * 10^9 * 5.4) / 59

r^2 = 8.2 * 10^8

r = 2.84 * 10^4 m

We can now find the Electric Potential by using:

V = kq/r

Hence,

V = (9 * 10^9 * 5.4) / (2.84 * 10^4)

V = 1.69 * 10^6 V

An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity of the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)

Answers

Answer:

45 W/m^2

Explanation:

Intensity of light, Io = 90 W/m^2

According to the law of Malus

$I=I_(0)Cos^(2)\theta$

The average value of Cos^θ is half

So, I = Io/2

I = 90 /2

I = 45 W/m^2

Final answer:

Unpolarized light, when passed through a polarizer, reduces its intensity by half. So, the intensity if the light that emerges from a vertical filter will be 45 W/m².

Explanation:

Given that the incident intensity of the unpolarized light is 90 W/m², when passed through a vertically oriented optical filter, the emerging light will be polarized and will have its intensity halved as it's the property of a polarizing filter to decrease the intensity of unpolarized light by a factor of 2. The formula used in this process is I = Io cos² θ. In the case of unpolarized light passing through a single polarizer, θ is 0. So, the formula simplifies to I = Io/2.

Therefore, the intensity of the light that emerges from the vertically oriented optical filter is: I = 90 W/m² / 2 = 45 W/m².

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wo charged spheres are 1.5 m apart and are exerting an electrostatic force (Fo) on each other. If the charge on each sphere decreases by a factor of 9, determine (in terms of Fo) how much electrostatic force each sphere will exert on the other.

Answers

Answer:

F0 / 81

Explanation:

Let the two charges by Q and q which are separated by d.

By use of coulomb's law

F0 = k Q q / d^2 ......(1)

Now the charges are decreased by factor of 9.

Q' = Q / 9

q' = q / 9 ......(2)

Now the Force is

F' = k Q' q' / d^2

F' = k (Q /9) (q / 9) / d^2

F' = k Q q / 81d^2

F' = F0 / 81

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 150-mL mark with 34.5°C acetone. After the flask is filled, the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32.0°C. The combination is then cooled back to 20.0°C. (The average volume expansion coefficient of acetone is 1.50 10-4(°C)−1.) (a) What is the volume of the acetone when it cools to 20.0°C?

Answers

Answer:149.73 ml

Explanation:

Given

$\beta \ of\ acetone=1.50* 10^(-4) ^(\circ)C^(-1)$

change in volume is given by

$\Delta V=V_(final)-V_(initial)$

$\Delta V=\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=\nu_(initial)+\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=150+150* 1.50* 10^(-4)\left [ 20-32\right ]$

$V_(final)=149.73 ml$

Final answer:

The volume of the acetone when it cools to 20.0°C is approximately 142.39 mL.

Explanation:

In order to determine the volume of the acetone when it cools to 20.0°C, we can use the equation for the volume change caused by a temperature change at constant pressure, known as Charles's law. Charles's law states that the volume of a gas is directly proportional to its temperature in Kelvin. We can use the formula V2 = V1 * (T2 / T1) to calculate the volume of the acetone at the lower temperature.

Given that the initial volume of the acetone is 150 mL at a temperature of 34.5°C, we need to convert this temperature to Kelvin by adding 273.15. Therefore, T1 = 34.5°C + 273.15 = 307.65 K.

Since the final temperature is 20.0°C, the final temperature in Kelvin will be T2 = 20.0°C + 273.15 = 293.15 K. We can now plug these values into the equation to find the volume of the acetone at the lower temperature: V2 = 150 mL * (293.15 K / 307.65 K) = 142.39 mL.

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"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

Answers

Answer:

0.5 lambda(wavelength)

Explanation:

We know that

The first harmonic for both side open ended pipe is

L= 1/2lambda

So L = 0.5*wavelength

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