A laser beam of wavelength λ=632.8 nm shines at normal incidence on the reflective side of a compact disc. The tracks of tiny pits in which information is coded onto the CD are 1.60 μm apart.For what nonzero angles of reflection (measured from the normal) will the intensity of light be maximum? Express your answer numerically. If there is more than one answer, enter each answer separated by a comma.

Answer:

D) resistance, resistivity

Explanation:

Resistance is a physical quantity that indicates the opposition of an object to conduct electricity, this quantity depends on different factors such as temperature, material, object length, among other things. The resistance of two objects of the same material may be different, because it depends on the specifications of the object.

On the other hand, resistivity is a more general quantity, since it is assigned to materials and depends only on the nature of the material and its temperature.

So the resistivity is related to the meterial rather than the object.

The answer is: Resistance is a property of an object while resistivity is a property of a material.

Answer:

a. 1.0 eV

Explanation:

Given that

Voltage difference ,ΔV = 1 V

From work power energy

Work =Change in the kinetic energy

We know that work on the charge W= q ΔV

For electron ,e= 1.6 x 10⁻¹⁹ C

q=e= 1 x 1.6 x 10⁻¹⁹ C

Change in the kinetic energy

ΔKE= q ΔV

Now by putting the values

ΔKE=  1 x 1.6 x 10⁻¹⁹   x 1   C.V

We can also say that

ΔKE=  1 e.V

Therefore the answer will be a.

a). 1.0 eV

Answer:

Explanation:

The body can be split into two parts

1 ) cylinder part of mass 7/8 x 64 = 56 kg of radius 20 cm

2 ) In the form of rod attached with cylinder having length 40 cm and mass

of 1/8 x 64 = 8kg

moment of inertia of cylinder

= 1/2 mr²

= .5 x 56 x ( 20 x 10⁻²)²

= 1.12 kg m²

moment of inertia of rods

= 1/3 ml²

= 1/3 x 8 x ( 40 x 10⁻²)²

= .4267 kg m²

Total MI = 1.5467 Kg m²

Answer:

Δ T = 2.28°C

Explanation:

given,

mass of marble = 100 Kg

height of fall = 200 m

acceleration due to gravity = 9.8 m/s²

C_marble = 860 J/(kg °C)

using conservation of energy

Potential energy = heat energy

  [tex]m g h = m C_{marble}\Delta T[/tex]

  [tex]g h =C_{marble}\Delta T[/tex]

  [tex]\Delta T= \dfrac{g h}{C_{marble}}[/tex]

  [tex]\Delta T= \dfrac{9.8 \times 200}{860}[/tex]

        Δ T = 2.28°C

The temperature change which occurs as a result of the fall, if we ignore energy gained or lost is 2.28°C

This is defined as the degree of hotness or coldness of a substance.

mass of marble = 100 Kg

height of fall = 200 m

acceleration due to gravity = 9.8 m/s²

C of marble = 860 J/(kg °C)

Using conservation of energy

mgh = mcΔT

ΔT = gh/c

= 9.8 × 200 / 860

 = 2.28°C

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Answer:

L = 0.44 [m]

Explanation:

Here we can use the Lorentz transformation related to length to solve it:

[tex]L=L_{0}\sqrt{1-\beta^{2}} [/tex]

Where:

L₀ is the length of the moving reference frame (penguin #1)

L is the length of the fixed reference frame (penguin #2)

β is the ratio between v and c

We know that v = 0.9c so we can find β.

[tex]\beta = \frac{0.9c}{c}=0.9[/tex]

[tex]L=1 [m]\sqrt {1-0.9^{2}} = 0.44 [m] [/tex]

Therefore, the length of the meter stick of #1 observed by #2 is 0.44 m.

I hope it helps you!

Answer:

Explanation:

Given

mass of object [tex]m=2 kg[/tex]

inclination [tex]\theta =50^{\circ}[/tex]

[tex]\mu _k=0.3[/tex]

[tex]\mu _s=0.4[/tex]

initial velocity [tex]u=3 m/s[/tex]

acceleration of block during upward motion

[tex]a=g\sin \theta -\mu _kg\cos \theta [/tex]

[tex]a=g(\sin 50-0.3\cos 50)[/tex]

[tex]a=5.617 m/s^2[/tex]

using relation

[tex]v^2-u^2=2a\cdot s[/tex]

where [tex]s=distance\ moved [/tex]

[tex]v=final\ velocity[/tex]

v=0 because block stopped after moving distance s

[tex]0-(3)^2=2\cdot (-5.617)\cdot s[/tex]

[tex]s=\frac{4.5}{5.617}[/tex]

[tex]s=0.801[/tex]

If block stopped after s m then force acting on block is

[tex]F=mg\sin \theta =[/tex]friction force [tex]f_r=\mu mg\cos \theta [/tex]

[tex]F>f_r[/tex] therefore block will slide back down to the bottom            

Final answer:

Calculations from the provided wave equation involve finding the period (T), horizontal wave travel, wave number and frequency, wave crest speed, and the maximum vertical displacement speed. These reveal wave motion characteristics including time intervals and speeds for complete wave patterns and individual points within a wave.

Explanation:

The student's question about a wave on a lake is one that involves understanding the characteristics and behavior of waves, which is a key concept in physics. Specifically, the student is asked to find the time for a complete wave pattern to pass a stationary observer, the horizontal distance traveled by a wave crest in that time, the wave number and frequency, the wave crest speed, and the maximum speed of a vertical displacement caused by the wave. Here's how we can calculate each of these:

Time for one complete wave pattern: The time taken for one complete wave to pass is the reciprocal of the frequency, known as the period (T). We use the angular frequency (5.05 s-1) provided in the equation y(x,t) = (3.30 cm) cos(0.400 cm-1x + 5.05 s-1t). We find T by taking 2π divided by the angular frequency.

Horizontal distance traveled: To find this, multiply the wave speed by the period (T).

Wave number and frequency: The wave number (k) is already provided as 0.400 cm-1, which can be converted to meters if needed. The frequency (f) is the angular frequency divided by 2π.

Wave crest speed: This can be found by dividing the angular frequency by the wave number.

Maximum speed of vertical displacement (cork floater): This is the amplitude times the angular frequency, which gives the maximum speed of any vertical movement due to the wave.

These calculations help understand the dynamics of wave motion and how waves interact with objects in their path like a fisherman's cork floater.

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2, the minimum tractive effort  = 220.3 lb.

To find the minimum tractive effort,

vehicle weight  is 2500 lb

drag coefficient is 0.35

frontal area 22 ft^2

vehicle speed is 60 mi/hr = 70 ft/sec

gradient - 5%

air density = 0.002045 slugs/ft^3

The force which exerted on a solid body moving in some relation to a fluid by the fluid's movement is known as a drag force.

Drag force is given as,

Fd = [tex]\frac{1}{2}[/tex]CdρAV²

= [tex]\frac{1}{2}[/tex] × 0.35 × 0.002045× 20× 88²

= 55.4 lb.

Force due to vehicle weight,

Fw = 0.01 ( 1+ υ/147)W

= 0.01 ( 1 + 88/147) 2500

= 39.965 lb.

Force due to gradient

Fg = W × g

= 2500 × 0.05

= 125 lb

Minimum tractive effort

F = Fd + Fw +Fg

 = 55.4 + 39.965 + 125

 = 220.365 lb

Thus, minimum tractive force can be found as 220.365 lb.

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Answer:

Part 1)

- 224.6 J

Part 2)

- 339.4 J

Explanation:

[tex]P[/tex] = Constant pressure acting on gas = 0.643 atm = (0.643) (1.013 x 10⁵) Pa = 0.651 x 10⁵ Pa

[tex]V_{i}[/tex] = initial volume = 5.46 L = 0.00546 m³

[tex]V_{f}[/tex] = final volume = 2.01 L = 0.00201 m³

Work done on the gas is given as

[tex]W = P (V_{f} - V_{i})\\W = (0.651\times10^{5}) ((0.00201) - (0.00546))\\W = - 224.6 J[/tex]

Part 2)

[tex]\Delta U[/tex] = Change in the internal energy

[tex]Q[/tex] = Heat energy escaped = - 564 J

Using First law of thermodynamics

[tex]Q = W + \Delta U\\- 564 = - 224.6 +  \Delta U\\ \Delta U = - 339.4 J[/tex]

Answer:

You have just driven across the Cold front.

Explanation:

Weather Front:

In Meteorology, weather fronts are simply boundaries between two air masses of different densities. There are four weather fronts and one of them is cold front and other fronts are warm front, stationary front and occluded front.

Cold Front:

In Meteorology, cold front is a boundary of cold air mass that is advancing under the warm air mass.

In our scenario, as the driver was moving in the west and he listened on radio, the next country on north is under tornado warning. So, when he passed the cold front, he experienced the conditions described.

To solve this problem it is necessary to apply the concepts related to the elastic potential energy from the simple harmonic movement.

Said mechanical energy can be expressed as

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

Where,

k = Spring Constant

A = Cross-sectional Area

From the angular movement we can relate the angular velocity as a function of the spring constant and the mass in order to find this variable:

[tex]\omega^2 = \frac{k}{m}[/tex]

[tex]k = m\omega^2\rightarrow \omega = 2\pi f [/tex]  for f equal to the frequency.

[tex]k = 1.53(2\pi 1.95)^2[/tex]

[tex]k = 229.44[/tex]

Finally the energy released would be

[tex]E = \frac{1}{2} (229.44)(0.075)^2[/tex]

[tex]E = 0.6453J \approx 0.646J[/tex]

Therefore the correct answer is B.

The correct answer is option A. The total mechanical energy of the system is [tex]\( 0.844 \, \text{J} \)[/tex]

To find the total mechanical energy of the system, we need to calculate both the elastic potential energy (due to the spring) and the gravitational potential energy (due to the height above the equilibrium position) at the maximum amplitude point.

 First, let's calculate the spring constant [tex]\( k \)[/tex] using the formula for the frequency[tex]\( k \)[/tex] of a mass-spring system:

[tex]\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \][/tex]

where \( m \) is the mass of the iron piece. Solving for [tex]\( k \)[/tex], we get:

[tex]\[ k = (2\pi f)^2 m \][/tex]

[tex]\[ k = (2\pi \times 1.95 \, \text{Hz})^2 \times 1.53 \, \text{kg} \][/tex]

[tex]\[ k \approx 244.73 \, \text{N/m} \][/tex]

Next, we calculate the elastic potential energy [tex]\( U_{\text{elastic}} \)[/tex] at the maximum amplitude [tex]\( A \)[/tex]:

[tex]\[ U_{\text{elastic}} = \frac{1}{2} k A^2 \][/tex]

[tex]\[ U_{\text{elastic}} = \frac{1}{2} \times 244.73 \[/tex], [tex]\text{N/m} \times (0.0750 \[/tex], [tex]\text{m})^2 \] \[ U_{\text{elastic}} \approx 0.683 \[/tex], [tex]\text{J} \][/tex]

Now, let's calculate the gravitational potential energy [tex]\( U_{\text{gravitational}} \)[/tex]at the maximum amplitude:

[tex]\[ U_{\text{gravitational}} = m g h \][/tex]

where [tex]\( g \)[/tex]is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \[/tex])) and [tex]\( h \)[/tex] is the height at maximum amplitude. Since the amplitude is given in centimeters, we convert it to meters:

[tex]\[ h = 7.50 \, \text{cm} = 0.0750 \, \text{m} \][/tex]

[tex]\[ U_{\text{gravitational}} = 1.53 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.0750 \, \text{m} \][/tex]

[tex]\[ U_{\text{gravitational}} \approx 1.130 \, \text{J} \][/tex]

The total potential energy [tex]\( U_{\text{total}} \)[/tex] at the maximum amplitude is the sum of the elastic and gravitational potential energies:

[tex]\[ U_{\text{total}} = U_{\text{elastic}} + U_{\text{gravitational}} \] \[ U_{\text{total}} \approx 0.683 \, \text{J} + 1.130 \, \text{J} \] \[ U_{\text{total}} \approx 1.813 \, \text{J} \][/tex]

However, we must consider that the total potential energy is zero at the equilibrium position. Therefore, the total mechanical energy [tex]\( E \)[/tex] of the system is equal to the total potential energy at the maximum amplitude:

[tex]\[ E = U_{\text{total}} \] \[ E \approx 1.813 \, \text{J} \][/tex]

Since the potential energy at the equilibrium position is zero, the total mechanical energy is simply the sum of the elastic and gravitational potential energies at the maximum amplitude, which is approximately [tex]\( 1.813 \, \text{J} \)[/tex]. However, the options given are in the range of [tex]\( 0.633 \, \text{J} \) to \( 0.955 \, \text{J} \)[/tex], which suggests that there might be a mistake in the calculation or interpretation of the problem.

Upon re-evaluating the calculation, it seems that the gravitational potential energy was incorrectly calculated. The correct calculation should consider that the gravitational potential energy is zero at the equilibrium position, not at the maximum amplitude. Therefore, we should only consider the change in gravitational potential energy from the equilibrium to the maximum amplitude, which is half the amplitude squared divided by the spring constant, as the mass will oscillate symmetrically around the equilibrium position.

The correct gravitational potential energy [tex]\( U_{\text{gravitational}} \)[/tex] at the maximum amplitude is:

[tex]\[ U_{\text{gravitational}} = \frac{1}{2} k \left(\frac{h}{2}\right)^2 \][/tex]

[tex]\[ U_{\text{gravitational}} = \frac{1}{2} \times 244.73 \, \text{N/m} \times \left(\frac{0.0750 \, \text{m}}{2}\right)^2 \][/tex]

[tex]\[ U_{\text{gravitational}} \approx 0.160 \, \text{J} \][/tex]

Now, the total mechanical energy [tex]\( E \)[/tex] of the system is:

[tex]\[ E = U_{\text{elastic}} + U_{\text{gravitational}} \] \[ E \approx 0.683 \, \text{J} + 0.160 \, \text{J} \] \[ E \approx 0.844 \, \text{J} \][/tex]

Answer

given,

mass of the car = 2510 Kg

angle of inclination = 10°

initial speed = v₁ = 20 m./s

skid length = 20 m

coefficient of friction = 0.5

Using conservation of energy

[tex]\Delta E = \Delta KE + \Delta U[/tex]

[tex]\Delta E = \dfrac{1}{2}mv^2 + m g h[/tex]

h = d sin θ

[tex]\Delta E = \dfrac{1}{2}mv^2- m g (dsin\theta)[/tex]

[tex]\Delta E = \dfrac{1}{2}\times 2510 \times 20^2- 2510\times 9.8 \times (20 sin10^0)[/tex]

[tex]\Delta E =416572.04\ J[/tex]

now, calculating the  magnitude of frictional force is equal to

E = F_f d

[tex]F_f = \dfrac{E}{d}[/tex]

[tex]F_f = \dfrac{416572.04}{20}[/tex]

[tex]F_f = 20828\ N[/tex]

Answer:

Explanation:

1 . When the Celsius temperature doubles, the Fahrenheit temperature doubles  - FALSE

A change of temperature from 300 F to 600 F changes celsius degree from 148 to 315 °C

2. When heat is added to a system, the temperature must rise.

- FALSE

When we add heat to a system of mixture of ice and water , temperature does not rise until whole of ice melts.

3. In a bimetallic strip of aluminum and brass which curls when heated, the aluminum is on the inside of the curve

- FALSE

Coefficient of linear thermal expansion of aluminium is more so it will be on the outer side.

4. An iron plate has a hole cut in its center. When the plate is heated, the hole gets smaller.

- FALSE

hole will be bigger. There will be photogenic expansion .

5.When the pendulum of a grandfather clock is heated, the clock runs more slowly.

TRUE

Time period is directly proportional to square root of length of pendulum.

when time period increases , clock becomes slow.

6.Cold objects do not radiate heat energy

- FALSE

Object radiates energy  at all temperature .

6 .

The statements are (1)False, (2)False, (3)True, (4)False, (5)True, (6)False

Here are the answers to your statements:

Answer:

.

Explanation:

Answer:

option (c)

Explanation:

mass of iron = 0.10 kg

mass of copper = 0.16 kg

rise in temperature, ΔT = 35°C

specific heat of iron = 450 J/kg°C

specific heat of copper = 390 J/kg°C

Heat by iron (H1) = mass of iron x specific heat of iron x ΔT

H1 = 0.10 x 450 x 35 = 1575 J

Heat by copper (H2) = mass of copper x specific heat of copper x ΔT

H1 = 0.16 x 390 x 35 = 2184 J

Total heat H = H1 + H2

H = 1575 + 2184 = 3759 J

by rounding off

H = 4000 J

Answer:b

Explanation:

Given

mass of iron Gear [tex]m_{iron}=0.1 kg[/tex]

mass of copper [tex]m_{cu}=0.16 kg[/tex]

specific heat of iron [tex]c_{iron}=450 J/kg-^{\circ}C[/tex]

specific heat of copper [tex]c_{cu}=390 J/kg-^{\circ}C[/tex]

heat gain by iron gear [tex]=m_{iron}c_{iron}\Delta T[/tex]

[tex]Q_1=0.1\times 450\times 35=1575 J[/tex]

heat gain by iron gear [tex]=m_{cu}c_{cu}\Delta T[/tex]

[tex]Q_2=0.16\times 390\times 35=2184 J[/tex]

Total heat [tex]Q_{net}=Q_1+Q_2[/tex]

[tex]Q_{net}=1575+2184=3759 J\approx 3800 J[/tex]

Answer:

22.1 m

Explanation:

[tex]v_{o}[/tex] = initial speed of ball = 14.3 m/s

[tex]\theta[/tex] = Angle of launch = 27°

Consider the motion of the ball  along the vertical direction.

[tex]v_{oy}[/tex] = initial speed of ball = [tex]v_{o} Sin\theta = 14.3 Sin27 = 6.5 ms^{-1}[/tex]

[tex]a_{y}[/tex] = acceleration due to gravity = - 9.8 ms⁻²

[tex]t[/tex]  = time of travel

[tex]y[/tex]  = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

[tex]y = v_{oy} t + (0.5) a_{y} t^{2} \\- 3.50 = (6.5) t + (0.5) (- 9.8) t^{2}\\- 3.50 = (6.5) t - 4.9 t^{2}\\t = 1.74 s[/tex]

Consider the motion of the ball along the horizontal direction.

[tex]v_{ox}[/tex] = initial speed of ball = [tex]v_{o} Cos\theta = 14.3 Cos27 = 12.7 ms^{-1}[/tex]

[tex]X[/tex]  = Horizontal distance traveled

[tex]t[/tex]  = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

[tex]X = v_{ox} t\\X = (12.7)(1.74)\\X = 22.1 m[/tex]

Answer:

b. 16.5°C

Explanation:

[tex]m_{c}[/tex] = mass of piece of copper = 80 g

[tex]c_{c}[/tex] = specific heat of piece of copper = 0.0920 cal/g°C

[tex]T_{ci}[/tex] = Initial temperature of piece of copper = 295 °C

[tex]m_{w}[/tex] = mass of water = 250 g

[tex]c_{w}[/tex] = specific heat of water = 1 cal/g°C

[tex]T_{wi}[/tex] = Initial temperature of piece of copper = 10 °C

[tex]m_{al}[/tex] = mass of calorimeter  = 300

[tex]c_{al}[/tex] = specific heat of calorimeter = 0.215 cal/g°C

[tex]T_{ali}[/tex] = Initial temperature of calorimeter = 10 °C

[tex]T[/tex] = Final equilibrium temperature

Using conservation of heat

Heat lost by piece of copper = heat gained by water + heat gained by calorimeter

[tex]m_{c} c_{c} (T_{ci} - T) = m_{w} c_{w} (T - T_{wi})+ m_{al} c_{al} (T - T_{ali})\\(80) (0.092) (295 - T) = (250) (1) (T - 10) + (300) (0.215) (T - 10)\\T = 16.5 C[/tex]

The final temperature of the system is 16.4 ⁰C.

The final temperature of the system is determined by applying the principle of conservation of energy as shown below;

Heat lost by piece of copper = heat gained by water + heat gained by calorimeter

mCc(Tc - T) = mCw(T - Tw) + mCl(T - Tl)

80 x 0.09 x (295 - T) = 250 x 1 x (T - 10) + 300 x 0.215 x (T - 10)

2124 - 7.2T = 250T - 2500 + 64.5T - 645

5269 = 321.7T

T = 5269/321.7

T = 16.4 ⁰C

Thus, the final temperature of the system is 16.4 ⁰C.

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Answer:

Frequency = 658 Hz

Explanation:

The third harmonics of A is,

[tex]f_{A}=(3)(440Hz)=1320Hz[/tex]

The second harmonics of E is,

[tex]f_{E}=(2)(659Hz)=1318Hz[/tex]

The difference in the two frequencies is,

delta_f = 1320 Hz - 1318 Hz = 2 Hz

The beat frequency between the third harmonic of A and the second harmonic of E is,

delta_f = [tex]3f_{A}-2f_{E}[/tex]

[tex]f_{E}=\frac{3f_{A}-delta_f}{2}[/tex]

We have calculate the frequency of the E string when she hears four beats per second, then

delat_f = 4 Hz

[tex]f_{E}=\frac{3(440Hz)-4Hz}{2}[/tex]

[tex]f_{E}=658Hz[/tex]

Hope this helps!

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

(a) Yes, this situation is possible. No, It is not possible in more than one way.

(b) The position of the wire is at a distance of 16.3 cm, in the direction left of the wire

(c) The magnitude of the current in wire 3 (I3) is 3.267 A and in a direction pointing in the downward direction.

(a)

The diagrammatic illustration of the information provided in the question can be seen in the image attached below.

From the image: suppose the forces on the wire as a result of the two other wires are equal and opposite, then the third wire will experience no net force. Hence, we can say Yes,  the situation is possible.

Given that:

It will not be possible to have more than one way, this is because the force acting on the other two wires will have opposite directions but they would not have the same equal magnitude.

The force per unit length existing in-between two current-carrying parallel wires can be determined by using the formula:

[tex]\mathbf{\dfrac{F}{l} = \dfrac{\mu_o I_1I_2}{2 \pi r} }[/tex]

where:

SInce F1 = F2, Then:

[tex]\mathbf{\dfrac{\mu_o\times I_1\times I_3\times l}{2 \pi r}= \dfrac{\mu_o\times I_2\times I_3\times l}{2 \pi (r+0.2)} }[/tex]

[tex]\mathbf{\dfrac{I_1}{r}= \dfrac{ I_2}{ (r+0.2)} }[/tex]

[tex]\mathbf{\dfrac{1.80}{r_1}= \dfrac{ 4.00}{ (r_1+0.2)} }[/tex]

1.80(r₁ + 0.2) = 4.00r₁

1.80r₁ +0.36 = 4.00 r₁

0.36 = 2.2r₁

r₁ = 0.36/2.2

r₁ = 16.3 cm in the direction left of the wire.

Again, we are to determine the magnitude of the current and its direction in wire 3 (I₃).

From the image, the forces of F21 = F23

∴

[tex]\mathbf{\dfrac{\mu_o \times I_1 \times I_2 \times l }{2 \pi r} = \dfrac{\mu_o \times I_2 \times I_3 \times l }{2 \pi r}}[/tex]

[tex]\mathbf{\dfrac{\mu_o \times 1.8 \times 4.0 \times l }{2 \pi \times 0.2} = \dfrac{\mu_o \times 4 \times I_3 \times l }{2 \pi \times 0.363}}[/tex]

Making I₃ the subject by equating both equations together, we have:

[tex]\mathbf{I_3 = \dfrac{1.8 \times 0.363}{0.2}}[/tex]

I₃ = 3.267 A and the current is pointing in the downward direction

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Answer:

In an elastic collision, the momentum and the kinetic energies are conserved.

Momentum:

[tex]\vec{P_i} = \vec{P_f}\\\vec{P}_1 + \vec{P}_2 = \vec{P}_1' + \vec{P}_2'\\m\vec{v_0} + 0 = m\vec{v_1}' + 3m\vec{v_2}}' \\v_0 = v_1 + 3v_2[/tex]

Kinetic energy:

[tex]K_i = K_f\\K_1 + K_2 = K_1' + K_2'\\\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}m{v_1'}^2 + \frac{1}{2}3m{v_2'}^2\\v_0^2 = {v_1'}^2 + 3{v_2'}^2[/tex]

We have two equations and two unknowns:

[tex]v_0 = v_1' + 3v_2'\\v_0^2 = {v_1'}^2 + 3{v_2'}^2\\\\3v_2' = v_0 - v_1'\\3{v_2'}^2 = {v_0}^2 - {v_1'}^2\\\\3{v_2'}^2 = (v_0 - v_1')(v_0 + v_1') = 3{v_2}'(v_1' + v_0)\\\\v_2' = v_1' + v_0\\3v_2' = v_0 - v_1'\\\\4v_2' = 2v_0\\\\v_2' = v_0/2\\v_1' = -v_0/2[/tex]

Explanation:

The first cart hits the second cart at rest and turns back with half its speed.

The second cart starts moving to the right with half the initial speed of the first cart.

Answer:

v1 = -v0/3 ,v2 = 2v0/3

Explanation:

Answer:

B) along the -z-axis.

Explanation:

Answer:

option (b)

Explanation:

direction of velocity is along + X axis

direction of force is along - Y axis

So, by use of the

[tex]\overrightarrow{F}=q\left ( \overrightarrow{v} \times \overrightarrow{B}\right )[/tex]

Where, B be the magnetic field

As charge of electron is negative

so the direction of magnetic field is along negative z axis .

The order of difficulty from the hardest to the easiest would be subject to the concept we have of Torque.

The Torque principle defines us that

[tex]\tau = Fd[/tex]

Where,

F = Force

d = Distance

As the distance increases, the force applied must be less to make the movement of the object so we have to

[tex]F \propto \frac{1}{d}[/tex]

Hence we have to be the distance inversely proportional to the force to turn the door the order would be:

As close to the hinges as possible (Hardest)> The center of the door> the edge farthest from the hinges (Easies)

1>2>3

The correct ranking of the difficulty of opening the door from hardest to easiest is as follows: 1. As close to the as possible,  2. The center of the door, 3. The edge farthest.

To understand the ranking, consider the physics of opening a door. The difficulty of opening a door is related to the torque required to rotate it around. Torque [tex](\(\tau\))[/tex] is calculated by the equation [tex]\(\tau = r \times F\)[/tex], where [tex]\(r\)[/tex] is the distance from the axis of rotation to the point where the force [tex](\(F\))[/tex] is applied.

1. As close possible: When you apply force close, the value of [tex]\(r\)[/tex] is smallest. This results in the smallest torque, meaning you have to exert more force to achieve the necessary torque to open the door, making it the hardest location to open the door.

2. The center of the door: Applying force at the center of the door increases the distance [tex]\(r\)[/tex]. This results in a larger torque for the same amount of force compared, but it is still not the most efficient point to open the door.

3. The edge farthest: The farthest edge provides the longest lever arm, meaning [tex]\(r\)[/tex] is at its maximum. This allows for the greatest torque for a given force, making it the easiest location to open the door. This is also why door handles are typically placed farthest.

Answer:

0.91125 km/s

Explanation:

[tex]v_1[/tex] = Velocity of planet initially = 54 km/s

[tex]r_1[/tex] = Distance from star = 0.54 AU

[tex]v_2[/tex] = Final velocity of planet

[tex]r_2[/tex] = Final distance from star = 32 AU

As the angular momentum of the system is conserved

[tex]mv_1r_1=mv_2r_2\\\Rightarrow v_1r_1=v_2r_2\\\Rightarrow v_2=\dfrac{v_1r_1}{r_2}\\\Rightarrow v_2=\dfrac{54\times 0.54}{32}\\\Rightarrow v_2=0.91125\ km/s[/tex]

When the exoplanet is at its farthest distance from the star the speed is 0.91125 km/s

Answer:

a. Δx = 2.59 cm

Explanation:

mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m

Using momentum conserved

mb * (0) + mp * vp = ( mb + mp ) * vf

vf = ( mp / mp + mb) * vp

¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²

Solve to Δx '

Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )

Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97  m /s) ²  / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]

Δx = 0.02599 m  ⇒ 2.59 cm

Answer

given,

[tex]y(x,t)=B cos[2\pi (\dfrac{x}{L} - \dfrac{t}{\tau})][/tex]

B = 6.40 mm ,  L = 26 cm ,    τ = 3.90 × 10⁻² s

general wave equation

  y = A cos (k x - ωt)

where A is the amplitude of the

a) Amplitude of the given wave

      B = 6.40 mm

b) Wavelength of the given wave

       λ = L

       λ = 26 cm

c)  wave frequency

     [tex]f = \dfrac{1}{\tau}[/tex]

     [tex]f = \dfrac{1}{3.9 \times 10^{-2}}[/tex]

            f = 25.64 Hz

d) speed of wave will be equal to

     v = f λ

     v = 25.64 x 0.26

     v = 6.67 m/s

e) direction of propagation will be along +ve x direction because sign of k x and ωt is same as general equation.

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

[tex]2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}[/tex]

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

[tex]2t = 2\lambda_{film}[/tex]

[tex]t = \lambda_{film}[/tex]

The index of refraction of soap is given, then

[tex]\lambda_{film} = \frac{\lambda_{vacuum}}{n}[/tex]

Combining the results of all steps we get

[tex]t = \frac{\lambda_{vacuum}}{n}[/tex]

Rearranging, we find

[tex]\lambda_{vacuum} = tn[/tex]

[tex]\lambda_{vacuum} = (278)(1.33)[/tex]

[tex]\lambda_{vacuum} = 369.74nm[/tex]

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

[tex]\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}[/tex]

Where

[tex]T_1 =[/tex]Temperature at inlet of turbine

[tex]T_2 =[/tex] Temperature at exit of turbine

[tex]P_1 =[/tex] Pressure at exit of turbine

[tex]P_2 =[/tex]Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

[tex]m_i = m_0 = m[/tex]

[tex]m(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W[/tex]

Where,

m = mass

[tex]m_i[/tex] = mass at inlet

[tex]m_0[/tex]= Mass at outlet

[tex]h_i[/tex] = Enthalpy at inlet

[tex]h_0[/tex] = Enthalpy at outlet

W = Work done

Q = Heat transferred

[tex]V_i[/tex] = Velocity at inlet

[tex]V_0[/tex]= Velocity at outlet

[tex]Z_i[/tex]= Height at inlet

[tex]Z_0[/tex]= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

[tex]h_i = h_0 + W[/tex]

[tex]W = h_i -h_0[/tex]

Using the relation T-P we can find the final temperature:

[tex]\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}[/tex]

[tex]\frac{T_2}{1400K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}[/tex]

[tex]T_2 = 725.126K[/tex]

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

So:

[tex]W = h_i -h_0[/tex]

[tex]W = C_p (T_1-T_2)[/tex]

[tex]W = 1.005(1400-725.126)[/tex]

[tex]W = 678.248kJ/Kg[/tex]

Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg

Final answer:

The maximum theoretical work that could be developed by the turbine, assuming ideal gas behavior, is approximately 0.3195 kJ per kg of air flow.

Explanation:

The maximum theoretical work that could be developed by the turbine can be calculated using the ideal gas law equation:

W = Cv(Tin - Tout)

where W is the work done per unit mass, Cv is the specific heat capacity at constant volume, Tin is the inlet temperature, and Tout is the outlet temperature.

First, we need to find the value of Cv for air. For ideal gases, the specific heat capacity at constant volume can be calculated using the equation:

Cv = (R/M)

where R is the gas constant and M is the molar mass of the gas.

In this case, we will use the molar mass of air, which is approximately 28.97 g/mol. The gas constant R is 8.314 J/(mol·K).

Substituting the values into the equation, we get:

Cv = (8.314 J/(mol·K)) / (28.97 g/mol)

Cv = 0.287 J/(g·K)

Now we can calculate the work:

W = (0.287 J/(g·K))(1400 K - 288 K)

W = 0.287 J/(g·K) x 1112 K

W = 319.544 J/g

Finally, we convert the work from Joules to kilojoules:

W = 319.544 J/g * (1 kJ/1000 J)

W ≈ 0.3195 kJ/g

The length of the neck that blood can still reach the brain depends on blood pressure, flow speed, and the cross-sectional area of capillaries. Using the given values, the approximate length of the neck is 9.8 meters.

The length of your neck that blood can still reach your brain depends on the blood pressure, flow speed, and the cross-sectional area of the capillaries. The blood enters the capillaries in the brain with a much smaller velocity of about 0.7 mm/s due to the large total cross-sectional area. To determine the length of the neck, we need to consider the time it takes for blood to travel from the heart to the brain. Using the given velocity, we can calculate that the time it takes for blood to reach the brain is approximately 14 seconds. Assuming a constant velocity, the length of the neck can be calculated using the formula: length = velocity x time, which gives us a result of 9.8 meters.

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The spring will stretch by 13.79 cm when 51.7 grams is hung on it.

To find the stretch of the spring when a weight of 51.7 grams is hung on it, we will use Hooke's Law. First, we need to convert the weight to Newtons. Since 1 g is equal to 0.0098 N, the weight in Newtons is 51.7 grams * 0.0098 N/g = 0.50546 N. Now we can rearrange Hooke's Law equation, F = -k * s, to solve for s, the stretch of the spring. Plugging in the values, we get 0.50546 N = -k * s. Rearranging further, we have s = 0.50546 N / -3.662 = -0.1379 m. Since the question asks for the answer in centimeters, we can convert -0.1379 m to centimeters by multiplying by 100. Therefore, the spring will stretch by 13.79 cm when 51.7 grams is hung on it.

Answer:

(a) maximum shear stress t=22.5mm

(b)maximum distortion energy t=19.48mm

Explanation:

Ф₁(sigma)=pd/2t

Ф₁=(5×10⁶×1.5)/2t

Ф₁=3.75×10⁶/t

Ф₂=pd/4t

Ф₂=1.875×10⁶/t

(a) According to maximum shear stress theory

|Ф₁|=ФY/1.5

3.75×10⁶/t=250×10⁶/1.5

t=22.5 mm

(b)According to distortion energy theory

Ф₁²+Ф₂²-(Ф₁Ф₂)=(ФY/1.5)²

(3.75×10⁶/t)²+(1.875×10⁶/t)-{(3.75×10⁶/t)(1.875×10⁶/t)}=(250×10⁶/1.5)²

t=19.48mm