X-rays are scattered from a target at an angle of 37.2° with the direction of the incident beam. Find the wavelength shift of the scattered x-rays.

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

[tex]F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}[/tex]

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

[tex]acceleration=\frac{30}{5}=6m/s^{2}[/tex]

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

[tex]v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds[/tex]

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

Answer:

Wavelength = 736.67 nm

Explanation:

Given

Energy of the photon = 2.70 × 10⁻¹⁹ J

Considering:

[tex]Energy=h\times frequency[/tex]

where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

So,  Formula for energy:

[tex]Energy=h\times \frac {c}{\lambda}[/tex]

Energy = 2.70 × 10⁻¹⁹ J

c = 3×10⁸ m/s

h = 6.63 x 10⁻³⁴ J.s

Thus, applying in the formula:

[tex]2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}[/tex]

Wavelength = 736.67 × 10⁻⁹ m

1 nm = 10⁻⁹ m

So,

Wavelength = 736.67 nm

Answer:

Explanation:

Area of plane = .35 x .7 = 0.245 m²

a) When plane is perpendicular to field ( plane is parallel to yz plane. )

Flux = field x area

3000 x .245 = 735 weber

b ) When plane is parallel to xy plane , the plane also becomes parallel to electric field so no flux will pass though the given plane.

Flux through the plane = 0

c ) Since normal to the plane makes 30 degree with x axis, it will also make 30 degree with direction of the field.

Flux through the plane

= electric field x area x cos 30

3000 x .245 x .866

636.51 weber.

Answer:

Navier Stokes equation

( 1 )  it is a partial differential equation that is describe the flow of incompressible fluids

Reynolds equation

(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication

Explanation:

Navier Stokes equation

( 1 )  it is a partial differential equation that is describe the flow of incompressible fluids

(2) Navier Stokes equation is used to model weather and ocean current and water flow in the pipe and air flow around wing

( 3) equation is

[tex]\nabla .\overrightarrow{v} = 0[/tex]   momentum equation

[tex]\rho \frac{d\overrightarrow{v}}{dt} = \nabla p + \rho \overrightarrow{g} + \mu \nabla ^2 v^2[/tex]

here [tex]\nabla p[/tex] is pressure gradient and [tex]\rho \overrightarrow{g}[/tex] is body force and [tex]\mu \nabla ^2 v^2[/tex] is diffusion term

and

Reynolds equation

(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication

(2) it is drive in 1886 from Navier Stokes law

(3) equation is attach

here

Answer:

Heat required to melt 1 lb of ice is 151.469 KJ

Explanation:

We have given mass of ice = 1 lb

We know that 1 lb = 0.4535 kg

Latent heat of fusion for ice =334 KJ/kg

Amount if heat for fusion of ice is given by

[tex]Q=mL[/tex], here m is mass of ice and L is latent heat of fusion

So heat [tex]Q=mL=0.4535\times 334=151.469kj[/tex]

So heat required to melt 1 lb of ice is equal to 151.469 KJ

Answer:

W = 289.70 kg

Explanation:

Given data:

Pressure in tank = 23 atm

Altitude 1000 ft

Air temperature  in tank T = 700 F

Volume of tank = 800 ft^3 = 22.654 m^3

from ideal gas equation we have

PV =n RT

Therefore number of mole inside the tank is

[tex]\frac{1}{n} = \frac{RT}{PV}[/tex]

              [tex] = \frac{8.206\times 10^{-5}} 644.261}{23\times 22.654}[/tex]

               [tex]= 1.02\times 10^{-4}[/tex]

               [tex]n = 10^4 mole[/tex]

we know that 1 mole of air weight is 28.97 g

therefore, tank air weight is [tex]W = 10^4\times 28.91 g = 289700 g[/tex]

               W = 289.70 kg

Answer:

[tex]Q=3.47\times 10^{-15}\ C[/tex]

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is [tex]q=1.6\times 10^{-19}\ C[/tex]

To find the net charge if there are n number of extra electrons is :

Q = n × q

[tex]Q=21749\times 1.6\times 10^{-19}\ C[/tex]

[tex]Q=3.47\times 10^{-15}\ C[/tex]

So, the net charge on the metal ball is [tex]3.47\times 10^{-15}\ C[/tex]. Hence, this is the required solution.

Final answer:

By applying the conservation of energy principle, accounting for the work done by the electric field as change in potential energy, we calculate the change in kinetic energy to find the speed of the charged particle at a different electric potential.

Explanation:

To calculate the speed of the particle at point B, we need to consider the change in electric potential energy and convert it to kinetic energy. The work done on the particle by the electric field is equal to the change in potential energy, which is given by W = q(VB - VA), where q is the charge of the particle, VA and VB are the electric potentials at points A and B respectively. Since no other forces act on the particle, the work done by the electric field causes a change in kinetic energy, which is defined by ½mv22 - ½mv12 = q(VB - VA), where m is the mass of the particle and v1 and v2 are the speeds at points A and B.

Rearranging the equation to solve for v2 gives us v2 = √(2q(VB - VA)/m + v12). Substituting in the given values, we can find the speed at point B.

Answer:

- 32.17 fts/s in the imperial system, -9.8 m/s in the SI

Explanation:

We know that acceleration its the derivative of velocity with respect to time, this is (in 1D):

[tex]a = \frac{dv}{dt}[/tex]

So, if we wanna know the change in velocity, we can take the integral:

[tex]v(t_f) - v(t_i) = \int\limits^{t_f}_{t_i} {\frac{dv}{dt} \, dt = \int\limits^{t_f}_{t_i} a \, dt[/tex]

Luckily for us, the acceleration in this problem is constant

[tex]a \ = \ - g[/tex]

the minus sign its necessary, as downward direction is negative. Now, for a interval of 1 second, we got:

[tex]t_f = t_i + 1 s[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  \int\limits^{t_i + 1 s}_{t_i} a \, dt[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  \int\limits^{t_i + 1 s}_{t_i} (-g ) \, dt[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  [-g t]^{t_i + 1 s}_{t_i} [/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g (t_i+1s) + g t_i [/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g t_i + -g 1s + g t_i[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g 1s [/tex]

taking g in the SI

[tex]g=9.8 \frac{m}{s^2}[/tex]

this is:

[tex]v(t_i + 1 s) - v(t_i) =  - 9.8 \frac{m}{s} [/tex]

or, in imperial units:

[tex]g=32.17 \frac{fts}{s^2}[/tex]

this is:

[tex]v(t_i + 1 s) - v(t_i) =  - 32.17 \frac{fts}{s} [/tex]

Answer:

Precise and Reproducible

Explanation:

For a system or a process to be useful a measurement standard for any physical quantity, it must be:

Answer:

precise

reproducible

Explanation:

Answer:

28.5 m/s

Explanation:

The speed has two orthogonal components, horizontal and vertical. To decompose the speed into these components we can use these trigonometric equations:

Vh = V * cos(a)

Vv = V * sin(a)

The forwards velocity is the horizontal component, so we use

Vh = V * cos(a)

Vh = 50 * cos(55)

Vh = 50 * 0.57

Vh = 28.5 m/s

For a mass-spring system with specified parameters, the damping constant is found to be 4.999 Ns/m. The amplitude reduces exponentially over time. The critical damping constant, defining the transition from underdamped to overdamped behavior, is 5 Ns/m.

This problem pertains to a system of mass springs undergoing damped oscillations. The damping constant (b) can be calculated using the formula b = 2*m*ωd. Here ωd signifies the damped frequency and is given by ωd = ω * (1 - γ/2), where ω is the natural frequency and γ is the damping ratio of 0.2% = 0.002. The natural frequency ω = sqrt(k/m) = sqrt(12.5/0.5) = 5 rad/s. Substituting these values in the equation gives b = 2*0.5*4.999 = 4.999 Ns/m.

The amplitude (A) of a damped oscillation decreases exponentially with time due to energy loss, described by A(t) = A0 * e^(-γωt/2), wherein A0 is the initial amplitude.

The critical damping constant (bc), describing the boundary between underdamped and overdamped systems, is equal to 2*sqrt(m*k) = 2*sqrt(0.5*12.5) = 5 Ns/m.

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Answer:

Explanation:

a) Hardness is a measure of the resistance of a material to permanent deformation (plastic) on its surface,

Hardness tests play an important role in material testing, quality control and component acceptance.

Hardness test are needed to be perform as a quality assurance procedure, to validate materials are according to the specific hardness required,

We depend on the data to verify the quality of the components to determine if a material has the necessary properties for its intended use.

Through the years, the establishment of increasingly productive and effective means of testing, has given way to new cutting-edge methods that perform and interpret hardness tests more effectively than ever. The result is a greater capacity and dependence on "letting the instrument do the work", contributing to substantial increases in performance and consistency and continuing to make hardness tests very useful in industrial and R&D applications.

b)

"The hardness test is a critical aspect of engineering practice for several reasons:

1. Material Performance: Hardness is a measure of a material's resistance to deformation, which is directly related to its wear resistance, strength, and durability. By performing hardness tests, engineers can predict how a material will perform under service conditions.

2. Quality Control: Hardness testing is used to ensure that materials meet the required specifications. It is a non-destructive (in some cases) or semi-destructive method to verify the quality of the material without compromising its integrity.

3. Material Selection: The results of hardness tests help engineers to select appropriate materials for different applications. A material that is too soft or too hard for a particular application can lead to premature failure or unnecessary costs.

4. Heat Treatment Verification: Heat treatment processes such as quenching, annealing, and tempering are used to alter the mechanical properties of materials. Hardness testing is used to confirm that the desired properties have been achieved after such treatments.

5. Failure Analysis: In the event of a material failure, hardness testing can provide valuable information about the cause of the failure. It can indicate if the material was too brittle or too soft for the intended application.

6. Comparative Analysis: Hardness tests provide a simple way to compare the properties of different materials or different batches of the same material.

7. Research and Development: In the development of new materials, hardness testing is an essential tool for characterizing materials and understanding their properties.

B) Two possible sources of error in hardness testing, other than parallax error, and their effects on the results are:

1. Indenter Geometry Errors: The geometry of the indenter (e.g., ball, cone, or pyramid) must be precise according to the test standard being used. Wear, damage, or incorrect shape of the indenter can lead to inaccurate measurements. For instance, a worn or blunt indenter will give lower hardness values than the actual hardness of the material.

2. Loading Rate and Duration Errors: The rate at which the load is applied and the duration for which it is held can significantly affect the hardness measurement. Some materials exhibit time-dependent plasticity (creep), which can affect the size of the indentation. If the loading rate or duration is not consistent with the standards or previous tests, the results may not be comparable or may not accurately represent the material's hardness. For example, applying the load too quickly may result in an underestimation of hardness due to the material's inability to fully resist the indenter.

To minimize these errors, it is essential to follow standardized testing procedures, maintain and calibrate equipment regularly, and ensure that the test conditions are consistent. Additionally, operators should be well-trained and aware of the potential sources of error to ensure the accuracy and reliability of the hardness test results."

Answer:

L =4166.66 N   and  D = 46296.29 N and T = 96047.34 N

Explanation:

given,

flight speed = 200 ft/s

turn rate = 5 deg/s

weight of aircraft = 50000 lb

L/D = 10

for equilibrium in the horizontal position

w - L cos ∅ - D sin ∅ = 0 .............(1)

D cos ∅ - L sin ∅ = 0................(2)

L/D = tan ∅ = 10

∅ = 84.29°

50000 - L cos  84.29°- D sin  84.29°= 0

D cos  84.29° - L sin 84.29° = 0

on solving the above equation we get

L =4166.66 N   and  D = 46296.29 N

thrust force calculation:

T = W sin ∅ + D

  = 50000×sin 84.29 + 46296.29

T = 96047.34 N

Final answer:

To find the time the football is in the air before hitting the ground, we can analyze the vertical motion using the given initial velocity and launch angle.

Explanation:

To find the time it takes for the football to hit the ground, we need to analyze the vertical motion of the football. We can use the formula:
t = (2 * vy) / g
where t is the time, vy is the vertical component of the initial velocity, and g is the acceleration due to gravity.

Given that the initial velocity is 22.0 m/s and the launch angle is 58.5°, we can find the vertical component of the velocity using the formula:
vy = v * sin(θ)
where v is the initial velocity and θ is the launch angle.

Using this information, we can calculate the time it takes for the football to hit the ground.

Answer:

a)a=2.63m/s^2

b)x=475.25m

Explanation:

To solve the first part of this problem we use the following equation

Vf=final speed=50m/s

Vo=initial speed=0

t=time

a=aceleration

a=(Vf-Vo)/t

a=(50-0)/19=2.63m/s^2

B )

For the second part of this problem we use the following equation

X=(Vf^2-Vo^2)/2a

X=(50^2-0^2)/(2*2.63)=475.25m

Answer:

[tex]v=0.04m/s\\[/tex]

[tex]s=-28.592m\\[/tex]

Explanation:

[tex]a = 3t-4[/tex]

[tex]v(t)=\int\limits^t_0 {a(t)} \, dt =3/2*t^{2}-4t+v_0\\[/tex]

if t=3.6s and initial velocity, v0,  is -5m/s

[tex]v=0.04m/s\\[/tex]

[tex]s(t)=\int\limits^t_0 {v(t)} \, dt =1/2*t^{3}-2t^{2}+v_0*t+s_0\\[/tex]

if t=3.6s and the initial displacement, s0, is -8m:

[tex]s=-28.592m\\[/tex]

Answer:

The car is traveling at [tex]8.1366\frac{m}{s}[/tex]

Explanation:

The known variables are the following:

[tex]V_{0} = 56 \frac{km}{h} = 15.56 \frac{m}{s}\\ D=25m\\ t=2.11s\\ V_{f}=?[/tex]

First, from the equation of motion we find the deceleration:

[tex]D=V_{0}*t+\frac{1}{2} a*t^{2} \\ a=\frac{2(D-V_{0})}{t^{2} } \\ a=3.5182\frac{m}{s^2}[/tex]

Then, with the equation for the speed:

[tex]V_{f}=V_{o}+a*t\\ V_{f}=8.1366\frac{m}{s}[/tex]

The speed of the car at impact is approximately 25.72 m/s.

To find the speed of the car at impact, we first need to calculate the deceleration. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

Given that the initial velocity (u) is 56.0 km/h, the distance (s) is 25.0 m, and the time taken (t) is 2.11 s, we can rearrange the equation to solve for a: a = (v - u) / t.

Plugging in the values, we have a = (0 - 56.0 km/h) / 2.11 s = -26.5 m/s². Since the car is decelerating, the acceleration is negative.

Now, to find the final velocity (v) at impact, we can use the equation v² = u² + 2as. Rearranging the equation, we have v² = u² + 2(-26.5 m/s²)(25.0 m) = u² - 2(26.5 m/s²)(25.0 m).

Plugging in the values, we have v² = (56.0 km/h)² - 2(26.5 m/s²)(25.0 m).

Converting the initial velocity (u) to m/s, we get u = 56.0 km/h * (1000 m/1 km) * (1 h/3600 s) = 15.6 m/s.

Substituting the values, we have v² = (15.6 m/s)² - 2(26.5 m/s²)(25.0 m) = 0.16 m²/s² - 26.5 m²/s² * 25.0 m = 0.16 m²/s² - 662.5 m²/s² = -662.34 m²/s².

Taking the square root of both sides, we get v ≈ √(-662.34 m²/s²) ≈ -25.72 m/s.

Since speed is a positive quantity, the speed of the car at impact is approximately 25.72 m/s.

Answer:

[tex]\eta = 0.411[/tex]

Explanation:

As we know that efficiency is defined as the ratio of output useful work and the input energy to the engine

So here we know that the

input energy = 441.3 kJ

energy rejected = 259.8 kJ

so we have

[tex]Q_1 - Q_2 = W[/tex]

[tex]W = 441.3 kJ - 259.8 kJ = 181.5 kJ[/tex]

now efficiency is defined as

[tex]\eta = \frac{W}{Q_1}[/tex]

[tex]\eta = \frac{181.5}{441.3}[/tex]

[tex]\eta = 0.411[/tex]

Answer:

3.85 nm

Explanation:

The volume of a sphere is:

V = 4/3 * π * r^3

The volume of a shell is the volume of the big sphere minus the volume of the small sphere

Vs = 4/3 * π * r2^3 - 4/3 * π * r1^3

Vs = 4/3 * π * (r2^3 - r1^3)

If the difference between the radii is 1 nm

r2 = r1 + 1 nm

Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)

Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)

Vs = 4/3 * π * (r1^3 + 3*r1^2 + 3*r1 + 1 - r1^3)

Vs = 4/3 * π * (3*r1^2 + 3*r1 + 1)

The volme of the shell is equall to the volume of the inner shell:

4/3 * π * (r1^3) = 4/3 * π * (3*r1^2 + 3*r1 + 1)

r1^3 = 3*r1^2 + 3*r1 + 1

0 = -r1^3 + 3*r1^2 + 3*r1 + 1

Solving this equation electronically:

r1 = 3.85 nm

Answer:

The force on X Fx=0 N

The force on Y Fy=-2.18 N

Explanation:

We have an array of charges, we will use the coulomb's formula to solve this:

[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]

but we first have to find the distance and the angule of the charge respect the charges on the Y axis:

[tex]r=\sqrt{(7*10^{-2}m)^2+ (3*10^{-2}m)^2} \\r=7.62cm=0.0762m[/tex]

we can notice that it is the same distance from both charges on Y axis.

we can find the angle with:

[tex]\alpha = arctg(\frac{7cm}{3cm})=66.80^o[/tex]

for the charge of 5µC [tex]\alpha =-66.80^o[/tex]

for the charge of -5µC [tex]\alpha =66.80^o[/tex]

the net force on the X axis will be:

[tex]F_{x5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(-66.80)\\F_{x5u}=0.465N[/tex]

and

[tex]F_{x(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(66.80)\\F_{x(-5u)}=-0.465N[/tex]

So the net force on X will be Zero.

for the force on Y we have:

[tex]F_{y5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(-66.80)\\F_{y5u}=-1.09N[/tex]

and

[tex]F_{y(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(66.80)\\F_{y(-5u)}=-1.09N[/tex]

Fy=[tex]F_{y5u}+F_{y(-5u)}[/tex]

So the net force on Y is Fy=-2.18N

To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. The force is calculated to be 5.12 N.

To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k * |q1 * q2| / r^2

Where F is the force, k is the electrostatic constant (k ≈ 8.99 * 10^9 N * m² / C²), q1 and q2 are the magnitudes of the charges, and r is the distance between them.

In this case, one charge is 5 µC and the other charge is -5 µC, with a distance of 14 cm between them (7 cm on the positive y-axis and 7 cm on the negative y-axis). The charge we're interested in is 2 µC at x = 3 cm. Plugging these values into the formula:

F = (8.99 * 10^9 N * m² / C²) * |(5 µC) * (2 µC)| / (0.14 m)²

F = 5.12 N

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Answer:

[tex]K_e_q=22.75878093\frac{N}{m}[/tex]

[tex]f=1.363684118Hz[/tex]

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

[tex]\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}[/tex]

Replacing the data provided:

[tex]\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}[/tex]

[tex]K_e_q=22.75878093\frac{N}{m}[/tex]

Finally, to calculate the frequency of oscillation we use this:

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

Replacing m and k:

[tex]f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz[/tex]

Answer:

Explanation:

Given

Velocity of helicopter relative to air 12.5 m/s to west

In vector form

[tex]V_{ha}=-12.5\hat{i}[/tex]

where [tex]V_h[/tex]= velocity of helicopter relative to ground

Also the velocity of air

[tex]V_a=4.55\hat{j}[/tex]

[tex]V_{ha}=V_h-V_a[/tex]

[tex]V_h=V_{ha}+V_a[/tex]

[tex]V_h=-12.5\hat{i}+4.55\hat{j}[/tex]

Speed relative to east[tex]=\sqrt{12.5^2+4.55^2}[/tex]

=13.302

For Direction

[tex]tan\theta =\frac{4.55}{-12.5}[/tex]

[tex]180-\theta =20[/tex]

[tex]\theta =160^{\circ}[/tex] relative to east

For 1 km travel it takes

[tex]t=\frac{1000}{13.302}=75.17 s[/tex]

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

[tex]V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s[/tex]

Part B. Using Kinematics expression for distance, time and initial velocity

[tex]y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s[/tex]

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.

The speed of the rock just before it hits the street is approximately 20.3 m/s, and the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.

Part A: To determine the speed of the rock just before it hits the street, we can use the concept of conservation of energy. When the rock is at the top of its trajectory, its potential energy is at its maximum and its kinetic energy is at its minimum. When the rock is just before hitting the street, its potential energy is at its minimum (zero) and its kinetic energy is at its maximum. Using the equations for potential energy and kinetic energy, we can calculate the speed of the rock:

Initial potential energy = final potential energy + final kinetic energy

mgh = 1/2m*v^2

where m is the mass of the rock, g is the acceleration due to gravity, h is the height of the building, and v is the final velocity of the rock.

Since the mass of the rock cancels out, we have:

gh = 1/2v^2

Plugging in the values, g = 9.8 m/s^2 and h = 21.0 m, we can solve for v:

v = sqrt(2gh)

v = sqrt(2*9.8*21.0)

v = sqrt(411.6)

v = 20.3 m/s

Therefore, the speed of the rock just before it hits the street is approximately 20.3 m/s.

Part B: To calculate the time elapsed from when the rock is thrown until it hits the street, we can use the equation for time of flight of a vertically thrown object:

t = 2v/g

where t is the time of flight, v is the initial upward velocity of the rock, and g is the acceleration due to gravity.

Plugging in the values, v = 27.0 m/s and g = 9.8 m/s^2, we can solve for t:

t = 2*27.0/9.8

t = 5.5 s

Therefore, the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.

Answer:

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

[tex]t_1 = \sqrt{\frac{2H}{g}}[/tex]

now the sound will come back to the observer in the time

[tex]t_2 = \frac{H}{v}[/tex]

so we will have

[tex]t_1 + t_2 = 2.42[/tex]

[tex]\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42[/tex]

so we have

[tex]\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42[/tex]

solve above equation for H

[tex]H = 26.8 m[/tex]

Part b)

If sound reflection part is ignored then in that case

[tex]H = \frac{1}{2}gt^2[/tex]

[tex]H = \frac{1}{2}(9.81)(2.42^2)[/tex]

[tex]H = 28.7 m[/tex]

so here percentage error in height calculation is given as

[tex]percentage = \frac{28.7 - 26.8}{26.8} \times 100[/tex]

[tex]percentage = 7.18 [/tex]

The janitor can determine the depth of the elevator shaft by timing the drop of the rock and using the displacement for free-falling objects formula, considering the time for the sound to travel up the shaft. The percent error if the sound travel time is ignored can be calculated by comparing the depth calculated with and without the sound travel time.

The janitor wants to determine how far it is from the point they dropped the rock to the bottom of the elevator shaft by measuring the time it takes for the rock to drop and hit the bottom. Here we can use the formula for the displacement of items in free-fall: d = 0.5gt², where d is the displacement (depth of the shaft), g is the acceleration due to gravity (9.8 m/s²), and t is the time it takes for the rock to fall (2.42s).

So, the depth of the shaft is d = 0.5 x 9.80 m/s² x (2.42 s)² = 28.63 m. However, we need to account for the time it takes for the sound to travel back up the shaft. The speed of sound is given as 336 m/s, and the time it takes to travel a certain distance is the distance divided by the speed, so in this case, it’s 28.63 m / 336 m/s = 0.085 s. Subtracting this from the total time gives us the true fall time of the rock, which we can plug back into the displacement formula to get the corrected depth of the shaft.

The percent error introduced if we ignore the travel time for the sound is then the depth obtained with sound travel time subtracted from the depth obtained without it, divided by the depth obtained with sound travel time, times 100.

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Answer and Explanation:

The inertial reference frame is one with constant velocity or non-accelerated frame of reference.

The value of acceleration and velocity change will vary in the two frames and will not be same.

As in case, we observe the acceleration and velocity of a moving train from the platform and the one observed in the train itself will be different.

In case of energy, it is dependent on the frame of reference but the energy change is independent of the frame of reference.

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = [tex]\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}[/tex]

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

The charges [tex]q_A[/tex] = 21.907nC and [tex]q_B[/tex] = 6.093nC are the charges on the beads.

We have to use Coulomb's Law and the principle of charge conservation.

1. Conservation of Charge:

  The total charge before and after they are touched must be the same. Initially, bead A has a charge of 28 nC, and bead B is neutral. After touching, let:

[tex]\[ q_A + q_B = 28 \, \text{nC} \][/tex]

2. Coulomb's Law:

  The force between two charges is given by Coulomb's Law:

[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]

Let's set up the equation:

[tex]\[4.8 \times 10^{-4} = 8.99 \times 10^9 \frac{q_A q_B}{(0.05)^2}\][/tex]

Rearrange to solve for [tex]\( q_A q_B \)[/tex]:

[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times (0.05)^2}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times 0.0025}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = \frac{1.2 \times 10^{-6}}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2\][/tex]

Now, we have two equations:

1. [tex]\( q_A + q_B = 28 \times 10^{-9} \, \text{C} \)[/tex]

2. [tex]\( q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2 \)[/tex]

To solve these, we can substitute [tex]\( q_B = 28 \times 10^{-9} \, \text{C} - q_A \)[/tex] into the second equation:

[tex]\[q_A \left( 28 \times 10^{-9} - q_A \right) = 1.3348 \times 10^{-16}\][/tex]

[tex]\[28 \times 10^{-9} q_A - q_A^2 = 1.3348 \times 10^{-16}\][/tex]

[tex]\[q_A^2 - 28 \times 10^{-9} q_A + 1.3348 \times 10^{-16} = 0\][/tex]

This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:

[tex]\[a = 1, \quad b = -28 \times 10^{-9}, \quad c = 1.3348 \times 10^{-16}\][/tex]

Solve using the quadratic formula [tex]\( q_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{(28 \times 10^{-9})^2 - 4 \times 1 \times 1.3348 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{7.84 \times 10^{-16} - 5.3392 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{2.5008 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm 1.5814 \times 10^{-8}}{2}\][/tex]

This gives us two solutions for [tex]\( q_A \)[/tex]:

1. [tex]\( q_A = \frac{28 \times 10^{-9} + 1.5814 \times 10^{-8}}{2} \\q_A = \frac{4.3814 \times 10^{-8}}{2} \\q_A = 2.1907 \times 10^{-8} \, \text{C} \\q_A = 21.907 \, \text{nC} \)[/tex]

2. [tex]\( q_A = \frac{28 \times 10^{-9} - 1.5814 \times 10^{-8}}{2} \\q_A = \frac{1.2186 \times 10^{-8}}{2} \\q_A = 6.093 \times 10^{-9} \, \text{C} \\q_A = 6.093 \, \text{nC} \)[/tex]

Since bead A has the greater charge, we take:

[tex]\[q_A = 21.907 \, \text{nC}\][/tex]

And the charge on bead B:

[tex]\[q_B = 28 \, \text{nC} - 21.907 \, \text{nC} = 6.093 \, \text{nC}\][/tex]

So, the charges on the beads are:

[tex]\[q_A = 21.907 \, \text{nC}, \quad q_B = 6.093 \, \text{nC}\][/tex]

Answer:

B. a conversion factor

Explanation:

The expression "1 in. = 2.54 cm" is called a conversion factor.

With this expression inches can be converted to centimeters.

Inversely the expression can also be used to convert centimeters to inches

By rearranging the equation we get

[tex]\frac{1}{2.54}\ inches=1\ cm\\\Rightarrow 1\ cm=0.394\ inches[/tex]

The statement which shows the equal nature of two different expressions is called an equation

Rule of thumb is a general approximation of a result of a test or experiment.

Hi your answer is B on edge

Answer:

a)[tex]v=6.19m/s[/tex]

b)[tex]v=5.51m/s[/tex]

c)[tex]a=3.3*10^{3}m/s^{2}[/tex]

d)[tex]x=5.78*10^{-3}m[/tex]

Explanation:

h1=195m

h2=1.55

a) Velocity just before the ball strikes the floor:

Conservation of the energy law

[tex]E_{o}=E_{f}[/tex]

[tex]E_{o}=mgh_{1}[/tex]

[tex]E_{f}=1/2*mv^{2}[/tex]

so:

[tex]v=\sqrt{2gh_{1}}=\sqrt{2*9.81*1.95}=6.19m/s[/tex]

b) Velocity just after the ball leaves the floor:

[tex]E_{o}=E_{f}[/tex]

[tex]E_{o}=1/2*mv^{2}[/tex]

[tex]E_{f}=mgh_{2}[/tex]

so:

[tex]v=\sqrt{2gh_{2}}=\sqrt{2*9.81*1.55}=5.51m/s[/tex]

c) Relation between Impulse, I, and momentum, p:

[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ (ma)*t=m(v_{f}-v{o})\\\\ a=\frac{ v_{f}-v{o}}{t}=\frac{ 5.51- (-6.19)}{3.5*10^{-3}}=3.3*10^{3}m/s^{2}[/tex]

d) The compression of the ball:

The time elapsed between the ball touching the ground and it is fully compressed, is half the time the ball is in contact with the ground.

[tex]t_{2}=t/2=3.5/2=1.75ms[/tex]

Kinematics equation:

[tex]x(t)=v_{o}t+1/2*a*t_{2}^{2}[/tex]

Vo is the velocity when the ball strike the floor, we found it at a) 6.19m/s.

a, is the acceleration found at c) but we should to use it with a negative sense, because its direction is negative a Vo, a=-3.3*10^3

So:

[tex]x=6.19*1.75*10^{-3}-1/2*3.3*10^{3}*(1.75*10^{-3})^2=5.78*10^{-3}m[/tex]

Answer:

The diameter is 0.378 ft.

Explanation:

Given that,

Mass of shot = 12 lb

Density of fresh water = 62.4 lb/ft

Specific gravity = 6.8

We need to calculate the volume of shot

[tex]V = \dfrac{4}{3}\pi r^3\ ft^3[/tex]

The density of shot is

Using formula of density

[tex]\rho = \dfrac{m}{V}[/tex]

Put the value into the formula

[tex]\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}[/tex]

We need to calculate the radius

Using formula of specific gravity

[tex]specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}[/tex]

Put the value into the formula

[tex]6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}[/tex]

[tex]r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}[/tex]

[tex]r^3=0.0067514[/tex]

[tex]r =(0.0067514)^{\frac{1}{3}}[/tex]

[tex]r=0.1890\ ft[/tex]

The diameter will be

[tex]d = 2\times r[/tex]

[tex]d =2\times0.1890[/tex]

[tex]d =0.378\ ft[/tex]

Hence, The diameter is 0.378 ft.