An explosion occurs at the end of a pier. The sound reaches the other end of the pier by traveling through three media: air, fresh water, and a slender metal handrail. The speeds of sound in air, water, and the handrail are 339, 1480, and 5060 m/s, respectively. The sound travels a distance of 141 m in each medium.

(a) After the first sound arrives, how much later does the second sound arrive?
(b) After the first sound arrives, how much later does the third sound arrive?

Answer:

c. The net force is vertically downward.

Explanation:

At the top of  the loop, the only external force that keeps the ball moving around the loop, is the centripetal force.

Now, this centripetal force, is not a " new" force, it's just the vector sum of the two external forces (neglecting friction) , that act simultaneously  on the marble, making it to change its speed, in magnitude and direction: the gravity force (which it is always downward), and the normal force(which is always perpendicular to the contact surface, preventing that the marble comes trough the surface), in this case between the marble and the track, which, at the top of the loop, points down too.

So, the net  force, exactly at the top of the loop, is vertically downward.

Answer:

111.657596

Explanation:

The expression of volume is given by

[tex]V=1001.93+111.5282+0.64698m^2[/tex]

Partially differentiating the term we get

[tex]\dfrac{\partial V}{\partial x}=\dfrac{\partial (1001.93+111.5282+0.64698m^2)}{\partial x}\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+2\times 0.64698m\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+1.29396m[/tex]

m = 0.100

[tex]\dfrac{\partial V}{\partial x}=111.5282+1.29396\times 0.100\\\Rightarrow \dfrac{\partial V}{\partial x}=111.657596[/tex]

The partial molar volume of glucose is 111.657596

Answer:

0.1040512455 N

[tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]

0.05925 N

[tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]

Explanation:

I = Current

B = Magnetic field

Separation between end points is

[tex]l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm[/tex]

Effective force is given by

[tex]F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N[/tex]

The force is 0.1040512455 N

[tex]tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}[/tex]

The angle the force makes is given by

[tex]\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]

The direction is [tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]

[tex]F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N[/tex]

The force is 0.05925 N

[tex]tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}[/tex]

[tex]\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]

The direction is [tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]

A. Magnitude of the force: [tex]\( 0.104 \, \text{N} \)[/tex]

B. Direction of the force: [tex]\( 323.8^\circ \)[/tex]

C. Magnitude of the force: [tex]\( 0.0593 \, \text{N} \)[/tex]

D. Direction of the force: [tex]\( 330^\circ \)[/tex]

Part A: Determine the magnitude of the magnetic force exerted by the external field on the wire

Given: Length of wire segments: [tex]\( L_1 = 40.0 \, \text{mm} = 0.040 \, \text{m} \) and \( L_2 = 55.0 \, \text{mm} = 0.055 \, \text{m} \)[/tex]

- Magnetic field, [tex]\( B = 0.300 \, \text{T} \)[/tex]

- Current, [tex]\( I = 5.10 \, \text{A} \)[/tex]

The force on a current-carrying wire in a magnetic field is given by [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \).[/tex]

For the first segment along the x-axis:

- [tex]\( \mathbf{L}_1 = 0.040 \, \text{m} \, \hat{i} \)[/tex]

[tex]\( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]

[tex]\[ \mathbf{F}_1 = I (\mathbf{L}_1 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.040 \, \hat{i} \times 0.300 \, \hat{k}) \][/tex]

[tex]\[ \mathbf{F}_1 = 5.10 \times 0.040 \times 0.300 \, \hat{j} \][/tex]

[tex]\[ \mathbf{F}_1 = 0.0612 \, \text{N} \, \hat{j} \][/tex]

For the second segment along the y-axis:

[tex]- \( \mathbf{L}_2 = 0.055 \, \text{m} \, \hat{j} \)[/tex]

[tex]- \( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]

[tex]\[ \mathbf{F}_2 = I (\mathbf{L}_2 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.055 \, \hat{j} \times 0.300 \, \hat{k}) \][/tex]

[tex]\[ \mathbf{F}_2 = 5.10 \times 0.055 \times (-0.300) \, \hat{i} \][/tex]

[tex]\[ \mathbf{F}_2 = -0.08415 \, \text{N} \, \hat{i} \][/tex]

The net force [tex]\(\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2\)[/tex]:

[tex]\[ \mathbf{F} = -0.08415 \, \hat{i} + 0.0612 \, \hat{j} \][/tex]

The magnitude of the force is:

[tex]\[ |\mathbf{F}| = \sqrt{(-0.08415)^2 + (0.0612)^2} \][/tex]

[tex]\[ |\mathbf{F}| = \sqrt{0.00708 + 0.00375} \][/tex]

[tex]\[ |\mathbf{F}| = \sqrt{0.01083} \][/tex]

[tex]\[ |\mathbf{F}| \approx 0.104 \, \text{N} \][/tex]

Part B: The direction of the force can be found using the angle [tex]\( \theta \)[/tex] with respect to the positive x-axis:

[tex]\[ \theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) \][/tex]

[tex]\[ \theta = \tan^{-1} \left( \frac{0.0612}{-0.08415} \right) \][/tex]

[tex]\[ \theta = \tan^{-1} \left( -0.727 \right) \][/tex]

[tex]\[ \theta \approx -36.2^\circ \][/tex]

Since this angle is measured counterclockwise from the positive x-axis, the direction clockwise is:

[tex]\[ 360^\circ - 36.2^\circ = 323.8^\circ \][/tex]

Part C: Determine the magnitude of the magnetic force exerted by the external field on the wire

For the wire directly from the origin to [tex]\( x = 20.0 \, \text{mm}, y = 35.0 \, \text{mm} \):[/tex]

- Length [tex]\( L = \sqrt{(20.0 \, \text{mm})^2 + (35.0 \, \text{mm})^2} = \sqrt{(0.020 \, \text{m})^2 + (0.035 \, \text{m})^2} \)[/tex]

- Length [tex]\( L = \sqrt{0.0004 + 0.001225} = \sqrt{0.001625} \)[/tex]- Length [tex]\( L \approx 0.0403 \, \text{m} \)[/tex]

[tex]\[ \mathbf{L} = 0.0403 \, \text{m} \][/tex]

Given current [tex]\( I = 4.90 \, \text{A} \)[/tex]:

[tex]\[ \mathbf{F} = I L B \sin(\theta) \][/tex]

[tex]\[ \theta = \angle \text{between } \mathbf{L} \text{ and } \mathbf{B} = 90^\circ \][/tex]

[tex]\[ \mathbf{F} = 4.90 \times 0.0403 \times 0.300 \][/tex]

[tex]\[ \mathbf{F} \approx 0.0593 \, \text{N} \][/tex]

Part D: - Current: [tex]\( 4.90 \, \text{A} \)[/tex]

- Magnetic field: [tex]\( 0.300 \, \text{T} \, \hat{k} \)[/tex]

- Displacement vector: [tex]\( \mathbf{L} = 0.020 \, \hat{i} + 0.035 \, \hat{j} \)[/tex]

- Force vector: [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \)[/tex]

[tex]\[ \mathbf{L} \times \mathbf{B} = (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]

[tex]\[ \mathbf{F} = 4.90 (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]

[tex]\[ \mathbf{F} = 4.90 \times (0.020 \times 0.300 \, \hat{j} - 0.035 \times 0.300 \, \hat{i}) \][/tex]

[tex]\[ \mathbf{F} = 4.90 \times (0.006 \, \hat{j} - 0.0105 \, \hat{i}) \][/tex]

[tex]\[ \mathbf{F} = 0.0294 \, \hat{j} - 0.05145 \, \hat{i} \][/tex]

The magnitude of the force:

[tex]\[ |\mathbf{F}| = \sqrt{(0.0294)^2 + (-0.05145)^2} \approx 0.0593 \, \text{N} \][/tex]

The angle with respect to the x-axis:

[tex]\[ \theta = \tan^{-1} \left( \frac{0.0294}{-0.05145} \right) = \tan^{-1} \left( -0.571 \right) \approx -30^\circ \][/tex]

The direction clockwise is:

[tex]\[ 360^\circ - 30^\circ = 330^\circ \][/tex]

Answer:

2.2 m/s to the left

Explanation:

Momentum is conserved, so:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

0 = (44 kg) (2.5 m/s) + (49 kg) v

v = -2.2 m/s

The canoe will move in the opposite direction of the woman with a velocity of approximately 2.24 m/s.

The velocity of the canoe after the woman jumps off can be determined by using the principle of conservation of momentum, which states that the total momentum before an event must equal the total momentum after the event when no external forces act on the system. In this scenario, the system consists of the woman and the canoe.

Before the woman jumps, the system is at rest, so its initial momentum is zero. When the woman jumps off the canoe to the right with a velocity of 2.5 m/s, by conservation of momentum, the canoe must move in the opposite direction to maintain the total momentum at zero.

The momentum possessed by the woman is given by the product of her mass and velocity (mw×vw).

Similarly, the momentum of the canoe is the product of its mass and its velocity in the opposite direction (mc×vc).

Momenta are equal in magnitude and opposite in direction:
mw ×vw = mc ×vc
Therefore, vc = (mw ×vw) / mc
Substituting the given values:

vc = (44 kg ×2.5 m/s) / 49 kg

= 2.24 m/s

The canoe will move to the left (opposite to the woman's direction) with a velocity of approximately 2.24 m/s.

Answer:

[tex]B=3A-5[/tex]

Explanation:

Variable Isolation

It's a common practice when dealing with equations that we have to isolate one variable in terms of other variables and/or constants. The isolation of the variable usually implies adding, subtracting, multiplying or dividing by constants. The following example shows how to isolate the A:

[tex]2A+3B=-4\\\\2A=-4-3B\\\\\displaystyle A=\frac{-4-3B}{2}[/tex]

We are required to find the equation where the variable has a coefficient of 1 and isolate it. The following equation fits into the description:

[tex]3A-B=5[/tex]

Isolating B:

[tex]B=3A-5[/tex]

We need to apply the definition of Young's double slit experiment. For fringe width of bright and dark fringe we have that

[tex]\beta = \frac{D\lambda}{d}[/tex]

Or expressed in terms of the wavelength we have that

[tex]\lambda = \frac{\beta d}{D}[/tex]

Where,

[tex]\lambda[/tex]= Wavelength

[tex]\beta[/tex]= Fringe width

d = Slit separation

D = Distance between slit and screen

From the ratios given in the equation, we have that as the wavelength decreases, the pattern determined for the diffraction pattern shrinks, which therefore causes all fringes to get narrower.

Final answer:

When the wavelength of monochromatic light decreases, the single-slit diffraction pattern shrinks with fringes getting narrower (option d). A decrease in slit width results in a wider diffraction pattern. More lines per centimeter on a diffraction grating cause bands to spread farther from the central maximum.

Explanation:

When a monochromatic light passes through a narrow slit, it forms a diffraction pattern due to the phenomenon of diffraction. The pattern consists of a series of bright and dark bands. The bright areas are known as maxima, while the dark areas are known as minima.

If the wavelength of the light decreases, the diffraction pattern shrinks, with all the fringes getting narrower (answer choice d). This is because the angle of diffraction is directly related to the wavelength, and as the wavelength becomes shorter, the angle at which light is bent decreases, causing the fringes to become narrower and the overall pattern to shrink.

Similarly, if the width of the slit producing a single-slit diffraction pattern is reduced, the pattern produced changes as well, with the bands spreading out and becoming wider. This is due to the inverse relationship between slit width and the angle of diffraction. A smaller slit width results in a larger spread of the diffraction pattern.

Lastly, if pure-wavelength light falls on a diffraction grating that has more lines per centimeter, the interference pattern will have bands that spread farther from the central maximum because a higher number of slits per unit area increases the diffractive effects.

Answer:True

Explanation:

The given statement is true because when we drive a car in a circle with constant speed , the car experiences the centripetal acceleration towards the center.

But acceleration is change in velocity of object in a given time, here direction of velocity is constantly changing to give rise to acceleration.

If the magnitude of velocity is changing with time then the car would have experiences the tangential acceleration .

Answer:

Option B is correct.

Explanation:

The resistance of a material or wire is represented by the formula below.

R=(ρL)/A

where

Resistance increases with increase in length and resistivity which means options C and D are wrong.

A=(π[tex]d^{2}[/tex])/4

An increase in diameter will result in a proportional increase in area

From the resistance formula, an increase in area will cause a reduction in the resistance of the material.Thus increased diameter of wire will lower the resistance of wire. Option B is correct

Have a great day

Answer

Mass of bullet (m) = .03 kg

Mass of wooden block M = 0.5 kg

Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height

finding the launch speed of bullet

Velocity of wooden block + bullet just after impact

= [\tex]\sqrt{2gh}[/tex]

=[\tex]\sqrt{2\times 9.8 \times 0.6}[/tex]

= 3.43 m/s

v₁ be the launch velocity

Applying law of conservation of momentum

0.03 x v₂ = 0.530 x 3.43

v₂ = 60.6 m /s

if v₁ be initial velocity

v₂² = v₁² + 2 g h

v₁² = v₂² - 2 gh

v₁² = 60.6 ² - 2 x (-9.8 )x 0.4

v₁ = 60.65 m /s this is launch speed

Answer:

77.08 C

Explanation:

[tex]m[/tex] = mass of the water = 500 g = 0.5 kg

[tex]c[/tex] = specific heat of water = 4186 J/(kg °C)

[tex]\Delta T[/tex] = Rate of change of temperature = 3 °C /min = (3/60 ) °C /s = 0.05  °C /s

[tex]k[/tex] = thermal conductivity of glass = 0.84

[tex]A[/tex] = Area of the element = 0.090 m²

[tex]t[/tex] = thickness of the element = 1.5 mm = 0.0015 m

[tex]T_{i}[/tex] = Temperature inside = 75 °C

[tex]T_{o}[/tex] = Temperature outside = ?

Using conservation of energy

Heat gained by water = Heat transferred through glass

[tex]m c \Delta T = \frac{kA(T_{o} - T_{i})}{t} \\(0.5) (4186 (0.05) = \frac{(0.84)(0.090)(T_{o} - 75)}{0.0015} \\104.65 = (50.4)(T_{o} - 75)\\T_{o} = 77.08 C[/tex]

Answer:

The energy of an electron in an isolated atom depends on b. n only.

Explanation:

The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.

The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.

The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates [tex]E_n[/tex], and for the case of an hydrogen atom we have:

[tex]E_n=-\cfrac{13.6}{n^2}\, eV[/tex]

Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.

Answer:

Part A:

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

Part B:

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

Part C:

a) If the distance to the screen is decreased the fringe spacing will decrease.

Part D:

The dot in the center of fringe E is [tex]920\ x\ 10^{-9} m[/tex] farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      [tex]\vartriangle y =\frac{m \lambda L}{d} [/tex]

                      [tex]m=0,\pm 1,\pm 2,\pm 3,.....[/tex]

        We simply replace the values in that equation :

                      [tex]\vartriangle r= m \lambda =2.\ 460\ nm[/tex]

                      [tex]\vartriangle r= 920\ x\ 10^{-9} m[/tex]

         The dot in the center of fringe E is [tex]920\ x\ 10^{-9}m[/tex] farther from the left slit than from the right slit.

By using the general equation for the double-slit maximums, we will get

The equation for the constructive interference in a double-slit experiment is given by:

[tex]y = \frac{m*\lambda*D}{d} [/tex]

Where:

Now let's answer:

A) If the wavelength is decreased, then the numerator is decreased, meaning the separation between consecutive fringes will also be decreased, so the correct option is a.

B) If d is decreased then the denominator decreases, meaning that the distance between consecutive fringes increases, so the correct option is b.

C) is the distance D is decreased, similar like in case A, the numerator decreases, meaning that the correct option is a again.

D) Sadly, as we do not know:

We can't answer this question.

What we should do here, is to compare the distance between the fringe and each slit, that distance will be the hypotenuse of a right triangle with one cathetus equal to D, and the other cathetus equal to y ± d/2, where each sign corresponds to each slit.

Then the difference in the distance will just be:

[tex]\sqrt{(y - d/2)^2 + D^2} - \sqrt{(y + d)^2 + D^2} [/tex]

If you want to learn more about the double-slit experiment, you can read:

https://brainly.com/question/13111431

Answer:

1. F = 45,458.17 N

2. P = 12,800,000 W

Explanation:

Part 1. The thrust force is the sum of the forces on the air and on the fuel.

For the air, 107 kg of air is accelerated from 281 m/s to 679 m/s in 1 second.

F = ma

F = (107 kg) (679 m/s − 281 m/s) / (1 s)

F = 42,586 N

For the fuel, 4.23 kg of fuel is accelerated from 0 m/s to 679 m/s in 1 second.

F = ma

F = (4.23 kg) (679 m/s − 0 m/s) / (1 s)

F = 2,872.17 N

So the thrust on the jet is:

F = 42,586 N + 2,872.17 N

F = 45,458.17 N

Rounded to three significant figures, the force is 45,500 N.

Part 2. Power = work / time, and work = force × distance, so:

Power = force × distance / time

Power = force × velocity

P = (45,458.17 N) (281 m/s)

P = 12,773,745.77 W

Rounded to three significant figures, the power is 12,800,000 W.

Answer:

a) [tex]\angle r_{ag}=38.79^{\circ}[/tex]

b) [tex]\angle r_{gw}=44.95^{\circ}[/tex]

c) not possible

Explanation:

Given:

angle of incidence on the air-glass interface, [tex]\angle i_{ag}=90-20=70^{\circ}[/tex]

refractive index of glass with respect to air, [tex]n_g=1.5[/tex]

refractive index of water with respect to air, [tex]n_a=1.33[/tex]

a)

For angle of refraction in glass we use Snell's law:

[tex]n_g=\frac{sin\ i_{ag}}{sin\ r{ag}}[/tex]

[tex]1.5=\frac{sin\ 70}{sin\ r_{ag}}[/tex]

[tex]\angle r_{ag}=38.79^{\circ}[/tex]

b)

Now we have angle of incident for glass-water interface, [tex]\angle i_{gw}=\angle r_{ag}=38.79^{\circ}[/tex]

And the refractive index of water with respect to glass:

[tex]n_{gw}=\frac{n_w}{n_g}[/tex]

[tex]n_{gw}=\frac{1.33}{1.5}[/tex]

[tex]n_{gw}=0.8867 [/tex]

Therefore, angle of refraction in the water:

[tex]n_{gw}=\frac{sin\ i_{gw}}{sin\ r_{gw}}[/tex]

[tex]0.8867 =\frac{sin\ 38.79}{sin\ r_{gw}}[/tex]

[tex]\angle r_{gw}=44.95^{\circ}[/tex]

c)

For total internal reflection through water the light must enter the glass-water interface at an angle greater than the critical angle.

So,

[tex]n_{gw}=\frac{sin\ i_{(gw)_c}}{sin\ 90}[/tex]

[tex]0.8867 =\frac{sin\ i_{(gw)_c}}{1}[/tex]

[tex]i_{(gw)_c}=62.46^{\circ}[/tex]

Now this angle will become angle of refraction for the air-glass interface.

Hence,

[tex]n_g=\frac{sin\ i_{ag}}{sin\ i_{(gw)_c}}[/tex]

[tex]1.5=\frac{sin\ i_{ag}}{0.8867 }[/tex]

[tex]sin\ i_{ag}=1.33005[/tex]

Since we do not have any angle for sine for which the value exceeds 1, therefore it is not possible that the light will reflect from the water.

To solve this problem we must use the perioricity equations given as a function of the mass and spring constant. Mathematically this can be expressed as:

[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]

m = mass

k = Spring constant

Re-arrange to find the mass we have

[tex]m = \frac{T^2k}{4\pi^2 }[/tex]

Replacing with our values we have that

[tex]m = \frac{2.4^2*10}{4\pi^2}[/tex]

[tex]m = 1.459kg[/tex]

D) Mass is independent of acceleration due to gravity (as you can see at the equation previously given) for this reason the mass suspended on mars is given as the same found. Therefore the mass would be

m = 1.459kg

Final answer:

The mass suspended from the spring that would result in a period of 2.4 s is 0.142 kg.

Explanation:

To find the mass suspended from the spring that would result in a period of 2.4 s, we can use the formula for the period of oscillation:

T = 2π √(m/k)

Where T is the period, m is the mass, and k is the force constant of the spring.

Let's rearrange the formula to solve for m:

m = (T^2 · k) / (4π^2)

Substituting the given values:

m = (2.4^2 · 10) / (4π^2)

m = 0.142 kg

Answer:

Explanation:

a.)

The magnitude field

[tex]B=\frac{\mu _0I_{enclosed}}{2\pi r}\\\\=\frac{\mu _0(J_ar^2)}{2\pi r}\\\\=\frac{1}{2}(\mu _0J_ar)\\\\(0.5)(4\pi \times 10^{-7})(20A/m^2)(87\times 10^{-3})\\\\=1.0933\times 10^{-6}T[/tex]

b.)

The displacement current

[tex]i_d=\epsilon_0A\frac{dE}{dt}[/tex]

then [tex]\frac{dE}{dt}=\frac{i_d}{\epsilon_0A}\\\\=\frac{J_d}{\epsilon_0}\\\\=\frac{20}{8.85\times 10^{-12}}\\\\=2.26\times 10^{12}V/ms[/tex]

Answer:

[tex]f_{e}[/tex]  = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

    M = M₀ [tex]m_{e}[/tex]

Where M₀ is the magnification of the objective and  [tex]m_{e}[/tex] is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

       [tex]m_{e}[/tex] = 25 /  [tex]f_{e}[/tex]

The objective is focused on the distances of the tube (L)

     M₀ = -L / f₀

Substituting

     M = - L/f₀    25/[tex]f_{e}[/tex]  

1) Let's look for the focal length of the eyepiece (faith)

     [tex]f_{e}[/tex]  = - L 25 / f₀ M

     M = 400X = -400

     [tex]f_{e}[/tex]  = - 12 25 /0.40 (-400)

     [tex]f_{e}[/tex]  = 1.875 cm

Let's approximate two significant figures

    [tex]f_{e}[/tex]  = 1.9 cm

Final answer:

The focal length of the eyepiece lens in the microscope is approximately -0.01 mm.

Explanation:

To find the focal length of the eyepiece lens, we can use the formula for the magnification of a compound microscope, which is given by:

M = -fobjective/feyepiece

Where M is the overall magnification, fobjective is the focal length of the objective lens, and feyepiece is the focal length of the eyepiece lens.

Given that the overall magnification is 400x and the focal length of the objective lens is 0.40 cm, we can rearrange the formula to solve for the focal length of the eyepiece lens:

feyepiece = -fobjective/M = -0.40 cm/400 = -0.001 cm = -0.01 mm.

Therefore, the focal length of the eyepiece lens is approximately -0.01 mm.

Answer:

2.112 MN

Explanation:

Force = mass × acceleration.  Each second, 264 kg of gas is accelerated from 0 to 8 km/s.

F = ma

F = m Δv / Δt

F = 264 kg × (8000 m/s − 0 m/s) / 1 s

F = 2,112,000 kg m/s²

F = 2.112 MN

Round as needed.

The thrust of the rocket, calculated using the formula - Thrust = mass rate of exhaust x exhaust velocity, is 2.112 Mega Newtons.

The thrust produced by a rocket is typically calculated using the formula: Thrust = mass rate of exhaust x exhaust velocity. In this scenario, the rocket is burning fuel at a rate of 264 kg/s (mass rate of exhaust), and the exhaust gas is ejected at a relative speed of 8 km/s.

However, to ensure the units are consistent, we have to convert the relative speed from km/s to m/s, which results in 8000 m/s (exhaust velocity).

Substituting these values into the formula, we get: Thrust = 264 kg/s x 8000 m/s = 2,112,000 N = 2.112 MN.

Therefore, the thrust of the rocket is 2.112 Mega Newtons.

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To solve this problem we will use the concepts related to the expression of energy for harmonic oscillator. From our given values we have that the period is equivalent to

[tex]T = 0.55s[/tex]

Therefore the frequency will be the inverse of the period and would be given as

[tex]f= \frac{1}{T}[/tex]

[tex]f = \frac{1}{0.55}[/tex]

[tex]f = 1.82s^{-1}[/tex]

The ground state energy of the pendulum is,

[tex]E = \frac{1}{2} hv[/tex]

[tex]E = \frac{1}{2}(6.626*10^{-34}J\cdot s)(1.82s^{-1})[/tex]

[tex]E = 6.03*10^{-34}J[/tex]

The ground state energy in eV,

[tex]E = 6.03 * 10^{-34}J(\frac{1eV}{1.6*10^{-19}J})[/tex]

[tex]E = 3.8*10^{-15}eV[/tex]

The energy difference between adjacent energy levels,

[tex]\Delta E = hv[/tex]

[tex]\Delta E = (6.626*10^{-34}J\cdot s)(1.82s)[/tex]

[tex]\Delta E = 12.1*10^{-34}J[/tex]

Answer:a

Explanation:

Given

Specific heat capacity of aluminium is twice that of iron.

[tex]c_{Al}=2c_{iron}[/tex]

Also mass of two blocks is equal

Rate of heat added is also same

[tex]\dot{Q_{Al}}=\dot{Q_{iron}}[/tex]

[tex]\frac{Q_{Al}}{t_{Al}}=\frac{Q_{iron}}{t_{iron}}[/tex]

[tex]\frac{mc_{Al}\Delta T}{t_{Al}}=\frac{mc_{iron}\Delta T}{t_{iron}}[/tex]

[tex]\frac{2c_{iron}}{t_{Al}}=\frac{c_{iron}}{t_{iron}}[/tex]

[tex]t_{Al}=2t_{iron}[/tex]

Thus Time taken by aluminium  block will be more                          

Answer:

3900000 m/s

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

v = Speed of the galaxy relative to the earth

Observed frequency is

[tex]f'=(1+0.013)f\\\Rightarrow \dfrac{f'}{f}=1.013[/tex]

Here the Doppler relation must be used.

So, observed frequency is given by

[tex]f'=f\dfrac{v+c}{c}\\\Rightarrow \dfrac{f'}{f}=\dfrac{v+c}{c}\\\Rightarrow v=\dfrac{f'c}{f}-c\\\Rightarrow v=(1.013\times 3\times 10^8)-3\times 10^8\\\Rightarrow v=3900000\ m/s[/tex]

The speed of the galaxy relative to the earth is 3900000 m/s

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

[tex]\eta = 1-\frac{T_L}{T_H}[/tex]

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

[tex]\eta = 1-\frac{T_L}{T_H}[/tex]

[tex]\eta = 1-\frac{293}{713}[/tex]

[tex]\eta = 0.589[/tex]

Therefore the maximum possible efficiency the car can have is 58.9%

Answer:

4.192 N

Explanation:

Step 1: Identify the given parameters

Velocity of airflow = 45m/s

air temperature = 20⁰

plate length and width = 1m and 0.5m respectively.

Step 2: calculate drag force due to shear stress, [tex]F_{s}[/tex]

[tex]F_{s} = C_{f} \frac{1}{2} (\rho{U_{o}}WL)[/tex]

Note: The density and kinematic viscosity of air at 20⁰ at 1 atm, is 1.2 kg/m³ and 1.5 X 10⁻⁵ N.s/m²

⇒The Reynolds number ([tex]R_{eL}[/tex]) based on the length of the plate is

[tex]R_{eL} =\frac{VXL}{U}[/tex]

[tex]R_{eL} =\frac{45X1}{1.5 X 10^{-5}}[/tex]

[tex]R_{eL}[/tex] = 3 X10⁶  (flow is turbulent, Re ≥ 500,000)

⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "tripped" side of the plate is

[tex]C_{f} = \frac{0.074}{(R_{e})^\frac{1}{5}}[/tex]

[tex]C_{f} = \frac{0.074}{(3 X10^6)^\frac{1}{5}}[/tex]

[tex]C_{f}[/tex] = 0.0038

⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "untripped" side of the plate is

[tex]C_{f} = \frac{0.523}{in^2(0.06XR_{e})} -\frac{1520}{R_{e}}[/tex]

[tex]C_{f} = \frac{0.523}{in^2(0.06 X 3X10^6)} -\frac{1520}{3X10^6}[/tex]

[tex]C_{f}[/tex] = 0.0031

The total drag force = [tex]\frac{1}{2}(1.2 X 45^2 X 1 X 0.5 (0.0038 +0.0031)[/tex]

The total drag force is 4.192 N

Answer:

[tex]2491 kg/m^3[/tex]

Explanation:

Suppose g = 9.8 m/s2. When the statue is suspended from the spring scale, the scale reads 28.4 N. This means the mass of that statue is:

[tex]m = N/g = 28.4 / 9.8 = 2.9 kg[/tex]

When the tub is lowered and submerged in water, the scale reads 17N. So the statue is subjected to a force that make the difference of 28.4 - 17 = 11.4N. This equals to the gravity force of water displaced.

[tex]\rho_wVg = 11.4[/tex]

Let water density [tex]\rho_w = 1000kg/m^3[/tex], we can calculate the volume of the water displaced, which is also the volume of the statue:

[tex]V = \frac{11.4}{g\rho_w} = \frac{11.4}{9.8*1000} = 0.00116 m^3[/tex]

The density of the statue is mass divided by its volume:

[tex]\rho = \frac{m}{V} = \frac{2.9}{0.00116} = 2491 kg/m^3[/tex]

This question involves the concepts of density, weight, volume, and buoyant force.

The density of the ceramic statue is "2495.5 kg/m³".

First, we will find out the mass of the statue:

[tex]W = mg\\m=\frac{w}{g}[/tex]

where,

W = hanging weight of statue = 28.4 N

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]m =\frac{28.4\ N}{9.81\ m/s^2}\\[/tex]

m = 2.9 kg

Now, we will find out the volume of the statue. The difference, in weight of the statue upon submerging, must be equal to the buoyant force applied by the water. This buoyant force is equal to the weight of the volume of water displaced, which is equal to the volume of the statue.

[tex]Difference\ in\ weight\ of\ statue=(Density\ of\ water)(Volume\ of\ Statue)g\\28.4\ N-17\ N=(1000\ kg/m^3)(V)(9.81\ m/s^2)\\\\V=\frac{11.4\ N}{(1000\ kg/m^3)(9.81\ m/s^2)}[/tex]

V = 1.16 x 10⁻³ m³

Now, the density of the ceramic is given as follows:

[tex]\rho = \frac{m}{V} = \frac{2.9\ kg}{1.16\ x\ 10^{-3}\ m^3}\\\\\rho=2495.5\ kg/m^3[/tex]

Learn more about buoyant force here:

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The attached picture illustrates the buoyant force.

Answer:

Explanation:

Given

Coefficient of drag [tex]C_d=1.1[/tex]

Reynolds number [tex]Re.no.=103 to 105[/tex]

velocity [tex]v=2 m/s [/tex]

[tex]F_d=3 N[/tex]

if velocity if 2v i.e. [tex]4 m/s [/tex]

[tex]F_d=\frac{1}{2}C_d\rho A v^2----1[/tex]

keeping other factors as constant

[tex]F'_d=\frac{1}{2}C_d\rho A (2v)^2----2[/tex]

dividing  1 and 2

[tex]\frac{F_d}{F'_d}=\frac{v^2}{2v^2}[/tex]

[tex]F'_d=4F_d[/tex]

[tex]F'_d=4\times 3=12 N[/tex]

Answer:

(a) 66.9%

(b) 147.14 kPa

Explanation:

Given:

Elevation of the water tank z = 15 m

Water volume flow rate V = 70 L/s = 0.07 m³/s

Input electric power consumption by the pump, Welec, in = 15.4  kW

Assuming there are no frictional losses in the pipes and changes in kinetic energy, the efficiency of the pump-motor will be;

                                     η =   ΔEmech ÷ Welec,

Where;

η is the overall efficiency

ΔEmech is the workdone to move the water pumped from the lake to a storage tank 15m above

Welec is the Input electric power consumption by the pump

Solving for ΔEmech ;

                         ΔEmech = mgh

                         mass, m = density × volume  

Density of water = 1000 kg/m³

                          m = 1000 kg/m³ ×  0.07 m³/s

                          m = 70 kg/s

            ∴ ΔEmech = 70 × 9.8 × 15

                               = 10.3 kW

Substituting the values of ΔEmech and Welec to calculate the overall efficiency

             η  =  (10.3 kW ÷ 15.4  kW) × 100 %

                   = 0.6688 × 100 %

                   = 66.88 %

                   = 66.9 %

The overall efficiency of the pump-motor unit is = 66.9 %

(b) The pressure difference between the inlet and the exit of the pump is calculated to be;

                                  Pressure = ΔEmech ÷ V  

                                                  = 10.3 ÷ 0.07

                                                  = 147.14 kPa

This question involves the concept of potential energy, pressure difference, and electrical work.

(a) Efficiency of pump-motor unit is "66.9 %".

(b) The pressure difference between the inlet and the exit of the pump is "147.15 KPa".

The efficiency of the pump-motor unit can be given by the following formula:

[tex]\eta = \frac{W_{P.E}}{W_{elect}}[/tex]

where,

Therefore,

[tex]\eta = \frac{\rho Vgh}{t\ W_{elect}}\\\\\eta=\frac{(1000\ kg/m^3)(0.07\ m^3/s)(9.81\ m/s^2)(15\ m)}{15400\ W}\\\\\eta =0.669 = 66.9\ \%[/tex]

The pressure difference between inlet and outlet of the pump can be found using the following equation:

[tex]\Delta P = \rho gh\\\\\Delta P = (1000\ kg/m^3)(9.81\ m/s^2)(15\ m)[/tex]

[tex]\Delta P = 147150\ Pa = 147.15\ KPa[/tex]

Learn more about the pressure difference here:

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Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

Expression for rate law for first order kinetics is given by:

[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = ?

t = time passed by the sample  = 90.3 min

a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

[tex]k=\frac{2.303}{90.3min}\log\frac{400}{100}[/tex]

[tex]k=1.54\times 10^{-2}\text{ min}^{-1}[/tex]

Now we have to calculate the half-life of substance, we use the formula :

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

[tex]1.54\times 10^{-2}\text{ min}^{-1}=\frac{0.693}{t_{1/2}}[/tex]

[tex]t_{1/2}=45min[/tex]

Therefore, the half-life of this substance will be, 45 minutes.

Answer:

0.49806

Explanation:

v = Velocity of wave

L = Length of tube

p denotes piccolo

f denotes flute

The fundamental frequency with both ends open is given by

[tex]f=\dfrac{v}{2L}\\\Rightarrow L=\dfrac{v}{2f}[/tex]

It can be seen that

[tex]L\propto \dfrac{1}{f}[/tex]

So,

[tex]\dfrac{L_p}{L_f}=\dfrac{f_f}{f_p}\\\Rightarrow \dfrac{L_p}{L_f}=\dfrac{257}{516} \\\Rightarrow \dfrac{L_p}{L_f}=0.49806[/tex]

The ratio of the piccolo's length to the flute's length is 0.49806

Answer:

Options A and D are correct

Explanation:

The thermal conductivity of a metal is the property of a metal to allow heat flow through it. conductivity is higher in conductors and low in insulators. Thermal conductivity is high in metals due to the metallic bonds that exist in metals and the presence of free electrons within the metal which allow easy flow of heat from one atom to another.From the problem the rod which contains freer electrons will allow more heat to flow easily hence have a higher thermal conductivity.

Thermal conductivity has the formula below;

[tex]k= \frac{QL}{AΔT}[/tex]

From the above equation we can see that thermal conductivity is inversely proportional to A and directly proportional to L. This mean the rod with less area will have a higher thermal conductivity and the rod with a higher length will have higher k. Hence option C i wrong and option D is correct.

For specific heat, its very much different from thermal conductivity. Specific heat is the ability of a material to hold heat while thermal conductivity is the ability of heat to flow through a material.

Answer:

       v₂ = 0.56 m / s

Explanation:

This exercise can be done using Bernoulli's equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Where points 1 and 2 are on the surface of the glass and the top of the straw

The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of ​​the straw the velocity of the surface of the vessel is almost zero v₁ = 0

The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m

We substitute in the equation

         [tex]P_{atm}[/tex] + ρ g y₁ = [tex]P_{atm}[/tex] + ½ ρ v₂² + ρ g y₂

         ½ v₂² = g (y₂-y₁)

        v₂ = √ 2 g (y₂-y₁)

Let's calculate

        v₂ = √ (2 9.8 1.6 10⁻²)

       v₂ = 0.56 m / s