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3 years ago

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A microscope has an objective lens with a focal length of 14.0mm . A small object is placed 0.80mm beyond the focal point of the

objective lens.Part A: At what distance from the objective lens does a real image of the object form? Express your answer to two significant figures and include the appropriate units.Part B: What is the magnification of the real image? Express your answer using two significant figures.Part C: If an eyepiece with a focal length of 2.5 cm is used, with a final image at infinity, what will be the overall angular magnification of the object? Express your answer using two significant figures.

1 answer:

Nikolay [14]3 years ago

3 0

Answer:

A. 260 mm

B. - 18

C. 175

Explanation:

A

The expression for the lens equation is

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{1}{14.0mm} = \frac{1}{14.80 mm} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{14.0} - \frac{1}{14.80}$

v = 259 mm

= 260 mm or 26 cm (to 2 s.f)

check the attached files for additional solution

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If you're driving 55 MPH and you suddenly need to stop, how many feet will you travel before the car comes to a stop

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ANSWER: 170 Feet
REASON: with good breaks and a dry road your car should stop and skip 170 feet, with perception and reaction time of stopping you should stop within 170 feet

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2 years ago

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

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Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

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2 years ago

Read 2 more answers

A research team developed a robot named Ellie. Ellie ran 1,000 meters for 200 seconds from the research building, rested for 100

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Answer:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

Explanation:

Attached you will find the plot of position vs time of Ellie´s movement.

The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.

So, the velocity is calculated as follows:

v = Δx / Δt

where

Δx = final position - initial position

Δt = elapsed time

1) The average velocity of Ellie while running is:

v = 1000 m - 0 m / 200 s = 5 m/s

While resting:

v = 0 m - 0 m / 100 s = 0 m/s

And while walking back:

v = 0 m - 1000 m / 1000 s = - 1 m/s

Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.

Comparing with the graphic, the velocity is the slope of the function position(t).

Then:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

2) The speed is the distance traveled over time:

Running speed = 1000 m / 200 s = 5m /s

Resting speed = 0 m / 100 s = 0 m/s

Walking speed = 1000 m/ 1000 s = 1 m/s

Then:

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

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2 years ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

$\texttt{ }$

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

<h3>Step B:</h3>

<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

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3 years ago

Which of the following statements is/are true? Check all that apply. A nonconservative force permits a two-way conversion betwee

saul85 [17]

Answer:

A conservative force permits a two-way conversion between kinetic and potential energies.

The work done by a nonconservative force depends on the path taken.

A potential energy function can be specified for a conservative force.

Explanation:

A conservative force is defined as a force whose work done does not depend on the path taken, but only on the initial and final position of motion.

This means that for a conservative force, it is possible to defined a potential energy function U which depends only on the position of the object. An example of conservative force is gravity: the gravitational potential energy of an object, in fact, depends only on its position in the field, not on the path taken.

This behaviour also implies that when an object moves from A to B and then back from B to A, the potential energy gained (or lost) moving from A to B is lost (or re-gained) when moving from B to A. This means that the total mechanical energy (sum of kinetic energy and potential energy) of the object is conserved, and therefore there is a constant conversion between potential and kinetic energy during the motion.

A non-conservative force instead does not show this properties, as the work done by it depends on the path taken, and therefore it is not possible to define a potential energy function. An example of non-conservative force is friction.

According to what we wrote above, therefore, the only correct statements are:

A conservative force permits a two-way conversion between kinetic and potential energies.

The work done by a nonconservative force depends on the path taken.

A potential energy function can be specified for a conservative force.

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