A fish swims 10 cm from the front wall of an aquarium that is 35cm wide. The front wall of the aquarium is glass with negligible thickness, but the back wall is a plane mirror. A person looks through the front wall and watches both the fish and its reflection in the mirror.Part A:What is the apparent distance from the front wall of the aquarium to the fish?Part B: What is the apparent distance from the front wall of the aquarium to the image of the fish in the mirror?

Answers

Answer 1

Answer:

El problema es un caso generalmente tipico en optica concerniente a Apparent depth vs real depth

We see the objects closer than their real depth to the surface. We see objects only if the rays coming from them reaches our eyes.

The equation is given by,

$D_a = (D_r)/(\eta)$

Where,

$D_a =$ Apparenth depth

$D_r =$ Real depth

$\eta =$ Refractive index of the medium of object

For water $\eta$ is equal to 1.33

I attach an image of the theory that could help clarify the measurements.

We have,

$D_a = (D_r)/(\eta)$

$D_a = (10)/(1.333)$

$D_a = 7.5cm$

Therefore the apparent distance between the front wall of the aquarium to the fish is 7.5cm

B) The distance between fish and mirror is given by,

$d=35-10= 25$

So we have that real distance from the front wall of to image of fish is

$dr=25+35=60cm$

Applying our equation we have that,

$D_a = (D_r)/(\eta)$

$D_a = (60)/(1.333)$

$D_a = 45.1cm$

Therefore the apparent distance from the front wall of the aquarium to the image of the fish is 45.1cm

Answer 2

Answer:

Final answer:

The apparent distance from the front of the aquarium to the fish is 10 cm, and the apparent distance from the front of the aquarium to the image of the fish in the mirror is 35 cm.

Explanation:

Part A: When observing the fish in the aquarium, the apparent distance from the front wall of the aquarium to the fish is simply the actual distance. This is because the observation is being made directly through the glass which has negligible thickness. Therefore, the apparent distance to the fish is 10 cm.

Part B: The image of the fish in the mirror will seem farther away than the fish itself. This is due to the fact that light reflects off the mirror and travels the distance of the aquarium twice. Hence, the total distance traveled by the light is the distance to the fish plus the distance from the fish to the mirror which is 35 cm - 10 cm = 25 cm. Thus, the apparent distance from the front wall of the aquarium to the image of the fish in the mirror is 10 cm (to the fish) + 25 cm (to the mirror) = 35 cm.

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The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer:

v = 4.1 m / s

Explanation:

Velocity is defined by the relation

v = $(dx)/(dt)$

we perform the derivative

v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

v = $(\Delta x)/(\Delta t)$

Δx = v Δdt + x₀

h= 4.1 t + 5.5

v = 4.1 m / s

x₀ = 5.5 m

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

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A 81.0 kg diver falls from rest into a swimming pool from a height of 4.70 m. It takes 1.84 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

Answers

Explanation:

The given data is as follows.

height (h) = 4.70 m, mass = 81.0 kg

t = 1.84 s

As formula to calculate the velocity is as follows.

$\nu$ = 2gh

= $2 * 9.8 m/s^(2) * 4.70 m$

= 92.12 $s^(2)$

As relation between force, time and velocity is as follows.

F = $(m * \nu)/(t)$

Hence, putting the given values into the above formula as follows.

F = $(m * \nu)/(t)$

= $(81.0 kg * 92.12 s^(2))/(1.84 s)$

= 4055.28 N

Thus, we can conclude that the magnitude of the average force exerted on the diver during that time is 4055.28 N.

According to the second law of thermodynamics, it is impossible for ____________. According to the second law of thermodynamics, it is impossible for ____________. heat energy to flow from a colder body to a hotter body an ideal heat engine to have the efficiency of 99% an ideal heat engine to have non-zero power. a physical process to yield more energy than what is put in

Answers

Answer:

It's impossible for an ideal heat engine to have non-zero power.

Explanation:

Option A is incomplete and so it's possible.

Option B is possible

Option D is related to the first lae and has nothing to do with the second law.

Hence, the correct option is C.

The ideal engine follows a reversible cycle albeit an infinitely slow one. If the work is being done at this infinitely slow rate, the power of such an engine is zero.

We can also stat the second law of thermodynamics in this manner;

It is impossible to construct a cyclical heat engine whose sole effect is the continuous transfer of heat energy from a colder object to a hotter one.

This statement is known as second form or Clausius statement of the second law.

Thus, it is possible to construct a machine in which a heat flow from a colder to a hotter object is accompanied by another process, such as work input.

Final answer:

According to the second law of thermodynamics, it is impossible for heat energy to flow from a colder body to a hotter body, for an ideal heat engine to have an efficiency of 99%, and for a physical process to yield more energy than what is put in.

Explanation:

According to the second law of thermodynamics, it is impossible for heat energy to flow from a colder body to a hotter body. This is because heat naturally flows from a region of higher temperature to a region of lower temperature. This principle is what allows us to effectively use heat for various purposes, such as in heat engines.

An ideal heat engine is a theoretical construct used to study the efficiency of engines. The second law of thermodynamics states that no heat engine can have an efficiency of 100%, so it is impossible for an ideal heat engine to have an efficiency of 99%. This is due to the losses in heat transfer and other thermodynamic processes.

The second law of thermodynamics also implies that in any physical process, the total energy cannot increase. It is impossible for a physical process to yield more energy than what is put in. This principle is central to understanding energy conservation and the limitations of energy conversion.

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A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The block is pulled 5.50 cm from equilibrium and released from rest. For the resulting oscillation, find the amplitude, angular frequency, frequency, and period. What is the maximum value of the block's velocity and acceleration?

Answers

Explanation:

It is given that,

Spring constant of the spring, k = 475 N/m

Mass of the block, m = 2.5 kg

Elongation in the spring from equilibrium, x = 5.5 cm

(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.

(b) Let $\omega$ is the angular frequency. It is given by :

$\omega=\sqrt{(k)/(m)}$

$\omega=\sqrt{(475)/(2.5)}$

$\omega=13.78\ rad/s$

(c) Let T is the period. It is given by :

$T=(2\pi)/(\omega)$

$T=(2\pi)/(13.78)$

T = 0.45 s

(d) Frequency,

$f=(1)/(T)$

$f=(1)/(0.45)$

f = 2.23 Hz

(e) Let v is the maximum value of the block's velocity. It is given by :

$v_(max)=\omega* A$

$v_(max)=13.78* 5.5* 10^(-2)$

$v_(max)=0.757\ m/s$

The value of acceleration is given by :

$a=\omega^2A$

$a=(13.78)^2* 5.5* 10^(-2)$

$a=10.44\ m/s^2$

Hence, this is the required solution.

Final answer:

The amplitude of the block-spring system is 0.055 m, with an angular frequency of 13.77 rad/s, and a frequency of approximately 2.19 Hz. The system has a period of approximately 0.46 s, with the maximum velocity being 0.76 m/s, and the maximum acceleration being 189 m/s².

Explanation:

In this block-spring system, we can determine the oscillation properties with the given parameters: mass (m = 2.50 kg), spring constant (k = 475 N/m), and displacement (x = 5.50 cm = 0.055 m).

Amplitude (A): This is the maximum displacement from the equilibrium position, in this case, it is 0.055 m, the distance from which the block is pulled and released.
Angular Frequency (ω): This can be calculated by the formula ω = √(k/m) which is equal to √(475/2.50) = 13.77 rad/s.
Frequency (f): It is given by the formula f = ω / 2π ≈ 2.19 Hz.
Period (T): The time for one complete cycle of the motion, calculated as T = 1/f ≈ 0.46 s.
Maximum value of velocity (v max): Calculated by the formula v = ω*A = 13.77 * 0.055 = 0.76 m/s.
Maximum value of acceleration (a max): Maximum acceleration occurs when the block is at the maximum displacement (amplitude), found with formula a = ω²*A = 189 m/s².

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Your classmate’s mass is 63 kg and the table weighs 500 N. Calculate the normal force on the table by the floor. Show your work!

Answers

Answer:

$F_N=1234.8N$

Explanation:

Hello.

In this case, since the normal force is opposite to the total present weight, we can compute it by considering the mass of the classmate with the gravity to compute its weight, and the weight of the table:

$F_N=63kg*9.8m/s^2+500N\n\nF_N=617.4N+500N\n\nF_N=1234.8N$

Best regards.

Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of the force that one particle exerts on the other?