Solve a problem.
An object 5 cm high stands in front of a converging lens at a distance of 20 cm and an image 2 cm high is formed. Determine the optical power and focal length of the lens.

Step 7: As we can see, there are numerous ways to position the three magnets to create the same pattern.

The pattern genuinely demonstrates repulsion. It indicates that the magnets' poles are the same since they are positioned that way.

A substance that generates a magnetic field and demonstrates magnetic qualities is known as a magnet. Iron, nickel, and cobalt are just a few of the metals that it can draw. Lodestone is an example of a naturally occurring magnetized mineral.

Magnets may also be manufactured artificially using techniques like magnetization. The north pole and the south pole are the two poles of a magnet.

Step 8: A different pattern of the three magnets can still produce a magnetic field.

There is fascination in the pattern. This indicates that the magnets' various poles attract.

Summing up:

1. It's possible that we can't tell whether is north or south.

2. We can actually distinguish between the ones that are north and south. This is seen in how they draw people to them.

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Answer:

16.53 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 18.0 m/s.

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

The maximum height reached by the ball can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 18² – (2 × 9.8 × h)

0 = 324 – 19.6h

Rearrange

19.6h = 324

Divide both side by 19.6

h = 324 / 19.6

h = 16.53 m

Therefore, the maximum height reached by the ball is 16.53 m

Answer:

2.63 microfarads

Explanation:

C = Q/V   =  2.5 x 10^-3 C / 950 V = 2.63 microfarads

A lens with an 11 cm focal length is 22 cm to the left of a 1.8 cm tall object. The image's height has decreased by 0.25 pixels when a second lens with a focal length of -5cm is placed 37cm to the right of the first lens.

Let's determine the distance at which the image is formed from the first lens since the image of the first lens serves as a virtual object for the second lens. Use the lens manufacturer's formula:

1/o + 1/i = 1/f, where I am the image distance and o is the object distance,

1/i = 1/f - 1/o = 1/(11cm) - 1/(22cm) = 0.045 1/cm, so

i = 22cm

the virtual object for the second lens is left at a distance of 37 cm - 22 cm, or 15 cm, from the second lens. Utilize the lens manufacturer's formula once more:

1/i = 1/f - 1/o (This time, the object distance is the separation between the second lens and the virtual object.)

1/i = 1/(-5cm) - 1/(15cm) = -0.26 1/cm, so

i = -15/4cm away from the second lens.

The sum of the two magnifications from the two lenses is the total magnification: m = m1m2, where m = -i/o, so

m = (-22 cm/22 cm)( 15/4cm/15cm) = -0.25, Consequently, the image's height is 0.25 pixels shorter than it was before.

It is the inverse of the system's optical power that determines how strongly an optical system diverges or converges light: the focal length. The convergence or divergence of light is indicated by a positive focal length, whilst the opposite is true for a negative focal length.

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A) To find the spring constant of the spring, we can use the conservation of energy. The spring constant of the spring is 1378.8 N/m.

B) The equilibrium point is at a height of 0 meters.

C)  The frequency of the oscillation is: f = 1/T ≈ 6.25 Hz

y(t) = 0.05 cos (2 π × 6.25 where y is in meters and t is in seconds.

The positive direction is from the equilibrium point.

a) To find the spring constant of the spring, we can use the conservation of energy. At the maximum height, the ball has no kinetic energy, so all the energy stored in the spring has been transferred to potential energy in the ball. The potential energy stored in a spring is given by:

PE = (1/2) k S^2

where k is the spring constant and S is the distance the spring is compressed. The potential energy stored in the spring must be equal to the potential energy of the ball at its maximum height. Using the given values, we can set up the equation:

(1/2) k S^2 = M g H

where M is the mass of the ball, g is the acceleration due to gravity, and H is the maximum height reached by the ball. Solving for k, we get:

k = 2 M g H / S^2

Substituting the given values, we get:

k = 2 × M × 9.81 m/s^2 × 0.35 m / (0.05 m)^2 = 1378.8 N/m

Therefore, the spring constant of the spring is 1378.8 N/m.

b) The equilibrium point of the ball when it is sitting on the spring with no forces other than gravity and the spring acting on it is the unscratched point of the spring. We can choose this point as the origin of our coordinate system, and take the upward direction as positive. Therefore, the equilibrium point is at a height of 0 meters.

c) When the ball is glued onto the spring and oscillates up and down, its motion can be described by a simple harmonic motion equation:

y(t) = A cos (ω t)

where y is the position of the ball, A is the amplitude of the oscillation, ω is the angular frequency, and t is the time. The amplitude of the oscillation is equal to the initial compression of the spring, which is 0.05 meters. The angular frequency is given by:

ω = 2 π f

where f is the frequency of the oscillation. The frequency of the oscillation is related to the period of the oscillation T by:

T = 1/f

The period of the oscillation can be found using the formula for the period of a simple harmonic motion:

T = 2 π √(m/k)

where m is the mass of the ball and k is the spring constant. Substituting the given values, we get:

T = 2 π √(0.1 kg / 1378.8 N/m) ≈ 0.16 s

Therefore, the frequency of the oscillation is:

f = 1/T ≈ 6.25 Hz

Substituting these values in the equation for the position of the ball, we get:

y(t) = 0.05 cos (2 π × 6.25 t)

where y is in meters and t is in seconds. The positive direction is upward from the equilibrium point.

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The object's final velocity, given the data is 10.5 rad/s

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

The following data were obtained from the question

The final velocity can be obtained as follow:

a = (v – u) / t

0.75 = (v – 1.5) / 12

Cross multiply

v – 1.5 = 0.75 × 12

v – 1.5 = 9

Collect like terms

v = 9 + 1.5

v = 10.5 rad/s

Thus, the final velocity of the object is 10.5 rad/s

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The caravan has an initial speed of

85 km/h × (1000 m/km) × (1/3600 h/s) ≈ 23.611 m/s

If it slows down at a constant rate, then it will cover a distance x such that

0² - (23.611 m/s)² = 2 (-1.3 m/s²) x

Solve for x :

x = -(23.611 m/s)² / (2 (-1.3 m/s²)) ≈ 214.417 m ≈ 210 m

Answer:

liquids

Explanation

To determine the time Pete arrives at work, we can start by calculating the total time he spends on his commute and gym routine:

Time spent driving to the gym = 12.5 miles ÷ average speed

We don't know Pete's average speed, so we cannot calculate this.

Time spent in the gym = 45 minutes

Time spent driving from the gym to work = 9 miles ÷ average speed

Again, we don't know Pete's average speed, so we cannot calculate this.

Total time spent on commute and gym routine = time spent driving to gym + time spent in gym + time spent driving from gym to work

= Unknown + 45 minutes + Unknown

Next, we can convert the total time to hours and minutes:

Total time = (Unknown + 45 minutes + Unknown) ÷ 60

= (Unknown + Unknown) ÷ 60 + 45/60

= (2Unknown) ÷ 60 + 0.75

= (Unknown) ÷ 30 + 0.75

We know that Pete needs to arrive at work by 9.00am, so we can set up an equation:

Arrival time = 7.30am + Total time

9.00am = 7.30am + (Unknown/30) + 0.75

Solving for Unknown:

1.5 hours = Unknown/30

Unknown = 45 minutes

Therefore, Pete will arrive at work at 8.15am.

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Answer:

M v^2 / R = centripetal force

For Red: M v^2 / R = 5 * 9^2 / 1.5 = 270

For Blue M v^2 / R = 270 = 25 * 1.8^2 / Rb

So Rb = 25 * 1.8^2 / 270 = .3 m

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

1 rev = 2(pi) rad  pi(rad) = 180 degrees

so 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s = 63.36 degrees

Explanation:

63.36 estimated to 63 so 63

The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.

1 rev = 2(pi) rad

So here  pi(rad) = 180 degrees

Now for 33 rpm it should be like

=  33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s

= 63.36 degrees

= 63 degrees

hence, The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.

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From the histogram given, there were at approximately 7 downloads between 3pm and 4pm . This can be derived by counting the rows in that time period.

A histogram is a graph that depicts the frequency distribution of a few data points from a single variable.

Histograms frequently divide data into "bins" or "range groups" and count the number of data points that belong to each of those bins.

Histograms are frequently used to depict the key properties of data distribution in a handy format. It is especially beneficial when working with huge data sets (more than 100 observations). It can aid in the detection of uncommon observations (outliers) or gaps in the data.2

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Answer:

Explanation:

Answer:

Explanation:

If we ASSUME that the mop has no acceleration, then the applied force must be acting parallel to the length of the handle.

a) 80.0cos40 = 61.3 N

b) 80.0sin40 =  51.4 N

The load lifted by the double pulley with the given efficiency and applied effort is determined as 30 N.

The amount of load lifted by the pulley is calculated by applying the formula for efficiency of a machine as follows;

E = L / E   x  100%

where;

The amount of load lifted by the pulley is calculated as;

60 = L / 50   x 100%

60 = 100L / 50

100L = 50 x 60

100 L = 3000

L = 3000 / 100

L = 30 N

Thus, the amount of load lifted by the pulley is 30 N.

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The complete question is below:

the efficiency of combine double pulley is 60%. how much load is lifted using 50n effort​?

Answer: 134 = 143 = 151 = 166 = 176

Hope this helps!!

Sorry if it's incorrect!!

:'(

Answer:

It bends.

Explanation:

Water is denser than water. This causes light that is passing from air to water to appear to bend towards the normal line near the water's surface.

Answer:

There is no movement

Explanation:

The reason why is because there is not enough froce to move it . PLEASE mark me brainliest!

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = \(\frac{\pi }{2}\) - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = \(\sqrt{2y_{o}/g}\)

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = \(\sqrt{k/m}\)

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - \(\sqrt{k/3}\) \(\sqrt{2y_{o}/g}\)

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = \(\frac{\pi }{2}\) - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Answer:

D

Explanation:

The gravitational pull of the planet it orbits holds a moon in its orbit.

the common name of sodium hydroxide is causyic soda

Given that:

For first hour- 40mph

For 3 hours - 70mph

30 minutes - 0 mph

after that- 70 mph

Here, the speed increases from 40 mph to 70 mph

So the graph here will be move upward.

For 30 min rest, the graph will be flat as distance is not changing with time.

After that it will again increase as it gains 70 mph speed.

Thus, the first graph is the correct one.

Answer:

60 kilometers per hr

Explanation:

12. The time taken for the journey is 400 s

13. The time taken for the train is 200 s

14. The time taken is 7 h

15. The time taken for the beetle is 12 s

16. The time taken for the journey is 0.0068 h

The time taken in each case as given by the question can be obtain as follow:

12. The time taken for the journey

Time taken = Distance / Speed

Time taken = 26000 / 65

Time taken = 400 s

13. The time taken for the train

Time taken = Distance / Speed

Time taken = 3200 / 16

Time taken = 200 s

14. The time taken to travel

Time taken = Distance / Speed

Time taken = 672 / 96

Time taken = 7 h

15. The time taken for the beetle

Time taken = Distance / Speed

Time taken = 1.08 / 0.09

Time taken = 12 s

16. The time taken for the journey

Time taken = Distance / Speed

Time taken = 35 / 5147

Time taken = 0.0068 h

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For two runners  running in a 500 meter dash and runner, the speeds of both runners  is mathematically given as

v1=25m/s

v2=20m/s

and the first runner is faster

Generally, the equation for the speed  is mathematically given as

v=d/t

Therefore, for first runner

v=500/20

v1=25m/s

For 2nd runner

v=500/25

v2=20m/s

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The CMB radiation's current temperature of 2.7 Kelvin is a result of the cooling and redshifting caused by the expansion of the Universe, which has been slowing down over time due to gravity.

The Cosmic Microwave Background (CMB) radiation is a remnant of the early hot and dense Universe. About 380,000 years after the Big Bang, the Universe cooled enough to allow the formation of neutral atoms, which made it transparent to radiation. This released the CMB radiation, which has been traveling through the Universe ever since, and its temperature has been cooling due to the expansion of the Universe. The CMB radiation is currently observed at a temperature of about 2.7 Kelvin, much cooler than the early Universe's temperature of 3000 Kelvin. This cooling is due to the expansion of the Universe, which causes the radiation to redshift to lower energies and longer wavelengths. This effect is known as the cosmological redshift. Moreover, the expansion of the Universe is not constant, but rather it is slowing down due to the gravitational pull of matter. This means that the early Universe's rate of expansion was faster than it is today, causing the CMB radiation to cool more rapidly in the past.

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Answer:

USB-6000

Explanation:

For this strange exercise, you must take into account two fundamental aspects of the data acquisition cards;

* The resolution of the cards, which for a 12-bit card is 2.24mV =2.24 10⁻³V , for which a card of this resolution is sufficient

    In your case, the amplitude is 50 mv, which is why it is much higher than the resolution of the card.

* The sampling rate or the frequency in which the card can acquire the data, the typical frequencies go from 200 Khz to the MHz, in general a card that has more than double the frequency to be measured should be selected.

In this case the frequency of your wave is

           f = 1 / T

           f = 1/100 10⁻³

            f = 10 Hz

therefore a low frequency card is suitable.

- The third point is the requirement that it has a USB interface, most of the card do not have it

If you check the National Instruments catalog, there are low-cost cards for example: USB-6000 of 12bits and 10KS / s the latter is the samples taken per second

\(\text{Voltage,}~ V = IR =33.72\times 32 = 1079.04~ \text{volt}\)