In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 52 m/s (terminal speed), that his mass (including gear) was 86 kg, and that the force on him from the snow was at the survivable limit of 1.2 ✕ 105 N.What is the minimum depth of snow that would have stopped him safely?

Answer:8.8 m/s

Explanation:

Given

mass of train [tex]m_1=0.6 kg[/tex]

velocity of Train [tex]v_1=20 m/s[/tex]

mass of another train [tex]m_2=1.5 kg[/tex]

Final velocity of both train [tex]v=12 m/s[/tex]

Let [tex]v_2[/tex] be the velocity of [tex]m_2[/tex] before collision

As external Force is zero therefore change in momentum is zero

conserving Momentum

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

[tex]0.6\times 20+1.5\times v_2=(0.6+1.5)\cdot 12[/tex]

[tex]1.5\times v_2=25.2-12[/tex]

[tex]v=\frac{13.2}{1.5}=8.8 m/s[/tex]    

The smallest detectable distance with the current setup is 0.98125cm. To make the robot travel 2m, the encoder needs to count roughly 204 times. If the smallest detectable change should be ±0.20cm, the encoder requires 80 lines per revolution.

The smallest detectable change in travel distance for the robot can be calculated based on the circumference of the wheel and the resolution of the rotary encoder. Given the diameter of the wheel is 10cm, the circumference can be calculated using the formula 2πr (r= radius of wheel), which results in 31.4cm. With 16 lines per revolution, this yields a distance per count (rise and fall) of 31.4cm/32 = 0.98125cm.

For the robot to travel 2m before making a turn, we will need to calculate the number of encoder counts. 2m equals 200cm, and if each count is representing a distance of 0.98125cm, then the number of encoder counts would be 200/0.98125, roughly equal to 204 counts.

If the goal is to reduce the smallest detectable distance to ±0.20cm, then we need to increase the lines per revolution of the encoder. If one line equates to 0.98125cm right now, that means we need roughly 5 times more lines on the encoder to achieve a resolution of 0.20cm. Therefore, the required lines per revolution would be 5*16 = 80 lines.

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Answer:

C. No work is done if the direction of motion is perpendicular to the force.

Explanation:

We know that work is the dot product of force and displacement.

Let's take the angle between the force and the displacement = θ

W= F . d cosθ

F=Force  , d=Displacement  

If θ = 0° then W= F.d

If θ = 90° then W= 0

So we can say that when force is perpendicular to the displacement then the work done by force will be zero.

Therefore the answer is C.

Answer:

2.79 m

Explanation:

Use the static equilibrium condition, net torque actin on the system is zero.

[tex]\sum \tau= 0[/tex]

[tex]T(Lsin\theta)- Mg\frac{L}{2}-mgx=0[/tex]

solve for the distance of the person from the wall x

[tex]x= \frac{TLsin\theta-Mg(L/2)}{Mg}[/tex]

now putting the values we get

[tex]x= \frac{1350\times4.25sin30-47\times9.81\times(4.25/2)}{69\times9.80}[/tex]

= 2.79 m

True

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Answer:

a. 1.12m/s

b. 0.34m/s

Explanation:

Let Vgi = the velocity of the girl relative to the ice

Let Vgp = the velocity of the girl relative to the plank

Let Vpi = the velocity of the plank relative to the ice

Vgi = Vgp + Vpi

From the question, Vgp = 1.46/s

So, Vgi = 1.46 + Vpi

Conservation of Momentum (Relative to the ice), we have

Initial Momentum = Final Momentum

Because the girl and the plank are at rest initially, the initial Momentum = 0

So, 0 = Mg * Vgi + Mp * Vpi

Where Mg = Mass of the girl = 45.8kg

Mo = Mass of the Plank = 151kg

Make Vpi the subject of the formula, we then have

-Mg*Vgi = Mp*Vpi ---------- Divide through by Mp

Vpi = -Mg * Vgi/Mp

Vpi =( -45.8 * Vgi)/151

Vpi = -45.8Vgi/151

Remember that, we have (Vgi = 1.46 + Vpi)

We then substitute the expression of Vpi in the above equation

That is;

Vgi = 1.46 + (-45.8Vgi/151) ------- Open the bracket

Vgi = 1.46 - 45.8Vgi/151 ----------- Multiply through by 151

151 * Vgi = 151 * 1.46 - 45.8Vgj

151Vgi = 220.46 - 45.8Vgi --------- Collect like terms

151Vgi + 45.8Vgi = 220.46

196.8Vgi = 220.46 --------------- Divide through by 196.8

Vgi = 220.46/196.6

Vgi = 1.1213631739572736

Vgi = 1.12 m/s (Approximated)

So, the velocity of the girl relative to the ice is 1.12m/s

b. Velocity of the plank, relative to the ice

We can solve this using (Vgi = 1.46 + Vpi)

All we need to do is substitute 1.12 for Vgi in the equation

So, we have

Vgi = 1.46 + Vpi becomes

1.12 = 1.46 + Vpi -------- Collect like terms

1.12 - 1.46 = Vpi

-0.34 = Vpi

So, the Velocity of the plank is 0.34m/s to the left

Answer: (a) 2.095 m/s (b) 0.635 m/s

Explanation: The initial momentum is = 0. Also, the final momentum is = 0. At the initial momentum she is at rest hence it is = 0. Whereas, the final momentum is = 0 because it has to equal the initial momentum.

This helps us to understand that when final momentum is = 0 then the girl's  direction would cancel the affect of the plank's momentum in the opposite direction.

Therefore,

Vg = Velocity of girl

Vp= Velocity of plank

(45.8)(Vg) + (151)(Vp) = 0

Girl's velocity relative to the plank is 1.46 m/s

Therefore, Vg - Vp = 1.46

Vg = 1.46 + Vp

(45.8)(1.46 + Vp) = 151Vp

66.868 + 45.8Vp = 151Vp

66.868 = 105.2Vp

Vp = 0.635 m/s

Vg = 1.46 + 0.635

Vg = 2.095

Final answer:

To determine the values requested - (a) the rms voltage across the resistor is 80.2 V, (b) the peak voltage of the source is 113.4 V, (c) the maximum current in the resistor is 8.02 A, and (d) the average power delivered to the resistor is 641 W.

Explanation:

In order to determine the values requested, we can use Ohm's law which states that V = IR, where V is the voltage, I is the current, and R is the resistance. Let's go through each part:

(a) To find the rms voltage across the resistor, we use the formula V = IR, where I is the ammeter reading and R is the resistance. Plugging in the values, we get V = (8.02 A) x (10.0 Ω) = 80.2 V.

(b) The peak voltage of the source can be found by multiplying the rms voltage by √2. So the peak voltage is (80.2 V) x √2 = 113.4 V.

(c) The maximum current in the resistor is equal to the ammeter reading, which is 8.02 A.

(d) The average power delivered to the resistor can be calculated using the formula P = IV, where I is the rms current and V is the rms voltage. Since the ammeter reading is the rms current, we can plug in the values to get P = (8.02 A) x (80.2 V) = 641 W.

Answer:

a) [tex] v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s[/tex]

b) [tex] v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s[/tex]

c) [tex] v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s[/tex]

d) [tex] v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s[/tex]

e) [tex] v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s[/tex]

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

[tex] KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)[/tex]           (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}} [/tex]

We can write the mass of a proton in MeV/c².

[tex] m_{p}=938.28 MeV/c^{2} [/tex]

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.04 [/tex]

[tex] v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s[/tex]

b) Linac (400 MeV)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.71 [/tex]

[tex] v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s[/tex]

c) Booster (8 GeV)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.994 [/tex]

[tex] v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s[/tex]

d) Main ring or injector (150 Gev)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.999 [/tex]

[tex] v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s[/tex]

e) Tevatron (1 TeV)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.9999 [/tex]

[tex] v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s[/tex]

Have a nice day!

The magnetic flux through an object like a bracelet can change a) because the bracelet is moving in the magnetic field, field direction, or if the bracelet moves through the field. Examples include loops moving into or rotating within a magnetic field.

The reason why the magnetic flux through the bracelet is changing can be due to various factors. These include a change in the magnitude of the magnetic field, a) a change in the field's direction relative to the bracelet, or the bracelet moving through the magnetic field. Magnetic flux changes when a loop moves into a magnetic field or rotates within it. Specifically, in part (a) as the loop moves into the field, and in part (b) as the loop rotates, thus changing the orientation.

It is also indicated that the change in magnetic flux through a loop or coil can result from the spinning of a magnet nearby, causing the magnetic field within the coil to change rapidly, as described with a rotating magnet affecting coil current due to the variation in the number and direction of magnetic field lines passing through it.

The correct option is c. The bracelet is moving in the magnetic field.

Magnetic flux through a surface, such as a bracelet, is defined as the product of the magnetic field (B), the area of the surface (A), and the cosine of the angle (θ) between the magnetic field lines and the normal to the surface. Mathematically, this is expressed as:

[tex]\[ \Phi_B = B \cdot A \cdot \cos(\theta) \][/tex]

 For the magnetic flux to change, one or more of the following must occur:

 1. The magnitude of the magnetic field (B) changes.

2. The area of the surface (A) changes.

3. The orientation of the surface with respect to the magnetic field changes, i.e., the angle (θ) changes.

4. The surface moves within the magnetic field, changing the amount of field lines passing through it.

 Given the options:

 a. The magnitude of the magnetic field is changing. - This would indeed change the magnetic flux, but it is not one of the given scenarios.

 b. The magnetic field is changing direction with respect to the bracelet. - This would also change the magnetic flux, as it would change the angle θ between the magnetic field lines and the normal to the surface of the bracelet. However, this option does not explicitly state that the direction of the magnetic field is changing; it only mentions the direction with respect to the bracelet, which could imply a change in orientation of the bracelet itself.

 c. The bracelet is moving in the magnetic field. - This option indicates that the bracelet is changing its position within the magnetic field. As the bracelet moves, the amount of magnetic field lines passing through it can change, even if the magnetic field itself is uniform and constant. This movement can alter the effective area through which the field lines pass (A) and/or the angle (θ), thus changing the magnetic flux.

Since the question asks for the reason why the magnetic flux through the bracelet is changing, and option c directly describes a scenario where the bracelet's movement within the field would cause a change in flux, it is the correct answer. The other options either do not apply to the given scenario or are not explicitly described as occurring.

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

[tex]\omega = \frac{v}{R}[/tex]

Where,

[tex]\omega =[/tex]Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

[tex]\alpha = \frac{\omega}{t}[/tex]

Where

[tex]\alpha =[/tex]Angular acceleration

[tex]\omega =[/tex] Angular velocity

t = Time

Our values are

[tex]v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})[/tex]

[tex]v = 16.67m/s[/tex]

[tex]r = 0.25m[/tex]

[tex]t=6s[/tex]

Replacing at the previous equation we have that the angular velocity is

[tex]\omega = \frac{v}{R}[/tex]

[tex]\omega = \frac{ 16.67}{0.25}[/tex]

[tex]\omega = 66.67rad/s[/tex]

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

[tex]\alpha = \frac{\omega}{t}[/tex]

[tex]\alpha = \frac{66.67}{6}[/tex]

[tex]\alpha = 11.11rad/s^2[/tex]

Therefore the angular acceleration of a point on the outer edge of the tires is [tex]11.11rad/s^2[/tex]

Using the Heisenberg Uncertainty Principle in quantum mechanics, which states that position and momentum of a particle cannot both be precisely measured at the same time, we can calculate the smallest box in which an electron can be confined with a specified maximum speed. The mass of the electron and the given speed are used to compute the uncertainty in momentum, which is then used in the Heisenberg inequality formula to find the size of the box.

This question can be addressed using the Heisenberg Uncertainty Principle, which in quantum mechanics, states that the position and the momentum of a particle cannot both be precisely measured at the same time. The more precisely we know the position (Δx), the less precisely we can know the velocity (Δv), and vice versa.

Considering an electron, which has a mass (m) of 9.11×10-31 kg, and wanting the speed to be no more than 13 m/s, we can use the Heisenberg inequality formula:

Δx* Δp ≥ ℏ/2,

where ℏ is the reduced Planck constant, and Δp is the uncertainty in momentum. Since momentum (p) is equal to the mass (m) times the velocity (v), we find the uncertainty in momentum (Δp) to be m* Δv which is (9.11×10-31 kg * 13 m/s). Now, we can solve for the smallest uncertainty in position (Δx), also known as the linear size of the box.

Using these calculations, we must first find Δp, then substitute into the above inequality and solve for Δx. This will give you the minimum size of the box in which the electron can be confined.

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The force on the wall can be calculated using the macroscopic balance for momentum. We need to calculate the mass flow rate of water and use the equation: mass flow rate = density × velocity × cross-sectional area of the nozzle. With the given values, we can find the force on the wall.

The force on the wall can be computed using the macroscopic balance for momentum. Since water discharges from a nozzle and travels horizontally, the initial momentum of the water is zero. The change in momentum of the water is equal to the force exerted on the wall. To calculate the force, we need to find the mass flow rate of the water. Assuming the water is an ideal fluid with a flat velocity profile, we can use the equation: mass flow rate = density × velocity × cross-sectional area of the nozzle. Plugging in the given values, we can then calculate the force on the wall.

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Final answer:

To calculate the force exerted on the wall by the water discharging from the nozzle, we use the mass flow rate and velocity of the water to find the momentum change. This calculation applies the macroscopic momentum balance principle and yields the force in Newtons.

Explanation:

The student is asking to calculate the force on a wall when water from a nozzle hits it horizontally. To find this force, we can use the macroscopic momentum balance principle which states that the rate of change of momentum of the fluid is equal to the sum of the external forces applied to the fluid.

As gravity and frictional effects are negligible, the only external force is the force on the wall exerted by the water. Key information given includes the diameter of the nozzle (12 mm) and the velocity of the water (6.0 m/s).

To calculate the force, we need to find the mass flow rate of the water, which can be determined by multiplying the density of water (ρ = 1000 kg/m³ at standard conditions) by the cross-sectional area of the nozzle (A) and the velocity (v) of the water.

The cross-sectional area, A, of the nozzle is A = π * (d/2)², where d is the diameter of the nozzle. Substituting the given values, the area A is π * (0.012/2)² m². The mass flow rate (ṭ) then is ρ * A * v.

With the mass flow rate known, the force (F) exerted on the wall can be calculated by taking the rate of change of momentum as ṭ * v, because the water's velocity is reduced to zero upon hitting the wall, thus the change in momentum is just the initial momentum. Therefore, F = ṭ * v.

After calculating the mass flow rate and applying the equation for force, the student will get the desired force in Newtons (N) on the wall.

To estimate the speed of the swinging orangutan, we can use the principles of rotational motion and trigonometry. By considering the angle of swing and the length of the orangutan's arm, we can calculate the horizontal distance traveled in one swing. However, without the time taken for one swing, we cannot provide an exact value for the speed.

To estimate the speed at which the orangutan is moving while swinging from branch to branch, we can use the principles of rotational motion and trigonometry. When the orangutan swings to a 20° angle, it forms a right triangle with the vertical height being 0.90 m and the hypotenuse being the length of the arm. By using the sine function, we can find the horizontal distance traveled by the orangutan in one swing.

Using the formula sin(20°) = horizontal distance / 0.90 m, we can rearrange the equation to find the horizontal distance traveled:

horizontal distance = 0.90 m * sin(20°)

Since the orangutan takes one swing immediately after another, the time taken for one swing is negligible. Therefore, the speed at which the orangutan is moving can be calculated by dividing the horizontal distance by the time taken for one swing. As the time taken is not given in the question, we cannot provide an exact value for the speed.

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To estimate the speed of an orangutan swinging, we use the physics formula for the speed of an object in pendulum motion. Under the given conditions of a 20° swing and arm length of 0.9 m, the orangutan's speed is approximately 2.54 m/s.

The behavior you're describing, where an orangutan swings like a pendulum beneath handholds, involves the study of pendulum motion, which is a concept in physics. Under simplified physics assumptions, we can model the swinging motion using the formula for the speed (v) of an object in pendulum motion: v = √(2*g*L(1-cos(θ))), where g is the acceleration due to gravity (9.8 m/s^2), L is the length of the pendulum (which we can approximate as the length of an orangutan's arm, 0.9 m), and θ is the angle swung through (20° or 0.349 radians after conversion).

Plugging the given values into the formula gives us: v = √(2*9.8*0.9(1-cos(0.349))), so the speed of an orangutan swinging would be approximately 2.54 m/s.

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Answer:

850N

Explanation:

The step by step is in the attachment.

Answer:C.

Explanation: the hoop is the right answer

Answer:

D) 50%

Explanation:

According to conservation of momentum:

[tex]m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}[/tex]

Assuming that the second taffy started at rest, both pieces have the same mass and that they combined after the collision, their final velocity is:

[tex]m v_{1i} = (m+m) v_{f}\\v_{f} = 0.5 v_{1i}[/tex]

The initial kinetic energy of the system is:

[tex]E_{ki} = \frac{m*v_{1i}^2}{2}[/tex]

Since the second taffy was not moving, it had no kinetic energy at first.

The initial kinetic energy of the system is:

[tex]E_{kf} = \frac{2m*v_{f}^2}{2}\\E_{kf} = \frac{2m*(0.5v_{1i})^2}{2}\\E_{kf} = \frac{0.5m*v_{1i}^2}{2}[/tex]

The percentage of kinetic energy that becomes heat is given by:

[tex]H=1 - \frac{E_{kf}}{E_{ki}}\\H=1 - \frac{\frac{0.5m*v_{1i}^2}{2}}{\frac{m*v_{1i}^2}{2}}\\\\H=1- 0.5 = 0.5[/tex]

THerefore, 50% of the kinetic energy becomes heat

In inelastic collisions, such as the described scenario where two objects stick together, 75% of the initial kinetic energy is lost to other forms, including heat.

The question involves a type of collision known as an inelastic collision, where two objects stick together post-collision. In these scenarios, while momentum is conserved, kinetic energy is not. The initial kinetic energy before the collision gets partly converted into other forms, such as heat.

Considering a scenario where a moving object collides with a stationary one of identical mass and they stick together, the resulting velocity of the combined mass is half of the original moving object's velocity. This result emerges from the conservation of momentum. Since kinetic energy is proportional to the square of the velocity, the kinetic energy of the combined object will be one-fourth of the initial kinetic energy. Consequently, 75% of the initial kinetic energy is converted into other forms of energy like heat, which answers the original question.

The speed of the car is 60 m/s

Explanation:

The speed of an object is a scalar quantity telling "how fast" the object is moving, regardless of its direction. It can be calculated as follows:

[tex]speed=\frac{d}{t}[/tex]

where

d is the distance covered

t is the time taken

For the car in this problem, we have

[tex]d=108 km =1.08\cdot 10^5 m[/tex] is the distance covered

[tex]t=30 min \cdot 60 =1800 s[/tex] is the time taken

Subsituting into the equation, we find the speed:

[tex]speed=\frac{1.08\cdot 10^5}{1800}=60 m/s[/tex]

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To solve the exercise it is necessary to apply the concepts given in Newton's second law and the equations of motion description.

Let's start by defining acceleration based on speed and time, that is

[tex]a = \frac{v}{t}[/tex]

On the other hand according to Newton's second law we have to

F=ma

where

m= Mass

a = Acceleration

Replacing the value of acceleration in this equation we have

[tex]F=m(\frac{v}{t})[/tex]

Substituting with our values we have

[tex]100N=(5Kg)\frac{v}{0.2}[/tex]

Re-arrange to find v

[tex]v=\frac{100*0.2}{5}[/tex]

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

Therefore the speed of the glider is 2m/s

Answer:

[tex]T = m*g*cos\theta[/tex]

Explanation:

Since tarzan moves in a circular trajectory we can sum all forces on the centripetal-axis:

[tex]T - m*g*cos\theta = m*a_c[/tex]

[tex]T - m*g*cos\theta = m*V^2/R[/tex]   Since the speed is zero:

[tex]T - m*g*cos\theta = 0[/tex]

[tex]T = m*g*cos\theta[/tex] Having Tarzan's mass we could calculate the module of the tension in the vine.

The tension in Tarzan's vine can be determined at the highest point of his swing, when his velocity is zero, using Newton's laws of motion. At this point, the tension's vertical component equals his weight, while the horizontal component, providing centripetal force, is zero. The equation T=mg/cosθ can be used to calculate the tension.

The subject of your question involves physics, particularly mechanics. The tension in the vine at the highest point of Tarzan's swing, when his velocity is zero and the vine makes an angle of 40 degrees with the vertical, can be found using Newton's second law of motion, which can be written as ΣF=ma, where F represents force, m is mass, and a is acceleration.

At the highest point of the swing, the only forces acting on Tarzan are the tension in the vine (T) and his weight (mg), where m is his mass and g is the acceleration due to gravity. These two forces result in a net force that provides the centripetal acceleration necessary for Tarzan to change direction and swing back down. We have:

So we can get the tension in vine (T) from the equation Tcosθ=mg. Rearranging, we have T=mg/cosθ.

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Answer:

22.02°C

Explanation:

P = Power output = 30 kW

Volume flow rate of water = [tex]1.49\times 10^{-4}\ m^3/s[/tex]

S = Specific heat of water = 4.186 kJ/mol°C

[tex]T_i[/tex] = Initial temperature = 10°C

[tex]T_f[/tex] = Final temperature

Total volume of water is

[tex]v=4\times 1.49\times 10^{-4}\\\Rightarrow v=5.96\times 10^{-4}\ m^3/s[/tex]

Mass flow rate is

[tex]m=\rho\times v\\\Rightarrow m=1000\times 5.96\times 10^{-4}\\\Rightarrow m=0.596\ kg/s[/tex]

Heat is given by

[tex]\frac{q}{t}=\frac{m}{t}S\Delta T\\\Rightarrow 30=0.596\times 4.186(T_f-10)\\\Rightarrow T_f=\frac{30}{0.596\times 4.186}+10\\\Rightarrow T_f=22.02\ ^{\circ}C[/tex]

The maximum possible temperature of the hot water that each showering resident receives is 22.02°C

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

[tex]\theta = 1.22\frac{\lambda}{d}[/tex]

Where,

[tex]\lambda[/tex]= Wavelength

d = Width of the slit

[tex]\theta[/tex]= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

[tex]r = l\theta[/tex]

Relacing [tex]\theta[/tex]

[tex]r = l(\frac{1.22\lambda}{d})[/tex]

[tex]r = 1.22\frac{l\lambda}{d}[/tex]

Replacing with our values we have that,

[tex]r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})[/tex]

[tex]r = 3680.91m[/tex]

Therefore the diameter of the beam on the moon is

[tex]d = 2r[/tex]

[tex]d = 2 * (3690.91)[/tex]

[tex]d = 7361.8285m[/tex]

Hence, the diameter of the beam when it reaches the moon is 7361.82m

Final answer:

Using diffraction formulas, it is estimated that a flashlight beam with an aperture of 7.0cm and an average wavelength of 550nm would have a diameter of approximately 3.7 km when it reaches the Moon.

Explanation:

To estimate the diameter of a flashlight beam when it reaches the Moon, we'll use the concept of diffraction spreading, as the beam is assumed to spread due to diffraction only. According to the diffraction formula for a circular aperture, θ = 1.22 λ/D. The angle θ is the angle of the spread, λ is the average wavelength of the light, and D is the diameter of the aperture through which the light passes.

Using the provided average wavelength of λ = 550nm and an aperture diameter of D = 7.0cm for the flashlight, we can calculate the minimum angular spread of the beam, θ.

θ = 1.22 × (550 x 10^-9 m) / (7.0 x 10^-2 m) = 0.0000095714 radians.

Then, to find out how broad the beam would be when it reaches the Moon, we use the angular spread to calculate the diameter of the beam at the distance to the Moon, which is 384,000 km (or 384 x 10^6 m):

Beam Diameter on the Moon = θ × Distance to the Moon = 0.0000095714 radians × 384 x 10^6 m = 3677.388 m.

Therefore, the flashlight's beam would have an estimated diameter of approximately 3.7 km when it reaches the Moon, considering only the diffraction spreading.

Answer:

(a) [tex]x_{n} = x_{n - 2} + x_{n - 4} + x_{n - 8}[/tex]

(b) 18 ways

Solution:

As per the question:

We need to find the no.of ways to go 'n' miles by foot:

From the question it is clear that, at each hours even miles can be traveled by a person, If the distance to be covered is odd, then there in no way 3 miles can be covered.

Thus defining recurrence as:

[tex]x_{o} = 1[/tex]

[tex]x_{n} = 0[/tex]

where

n: negative or odd

Then

[tex]x_{n} = x_{n - 2} + x_{n - 4} + x_{n - 8}[/tex]

(b) No. of ways to travel 12 miles:

[tex]x_{o} = 1[/tex]

[tex]x_{2} = 1 + 0 = 1[/tex]

[tex]x_{4} = x_{2} + x_{o} + 0 = 1 + 1 = 2[/tex]

[tex]x_{6} = x_{4} + x_{2} + 0 = 2 + 1 + 0 = 3[/tex]

[tex]x_{8} = x_{6} + x_{4} + 0 = 3 + 2 = 6[/tex]

[tex]x_{10} = x_{8} + x_{6} + x_{2} = 6 + 3 + 1 = 10[/tex]

[tex]x_{12} = x_{10} + x_{8} + x_{4} = 10 + 6 + 2 = 18[/tex]

Thus clearly there are 18 ways to travel 12 miles

The recurrence relation for the number of ways to go n miles by foot at different speeds is An = An-2 + An-4 + An-8. To find the number of ways to go 12 miles, we can compute the values of An using the recurrence relation.

To find a recurrence relation for the number of ways to go n miles by foot walking at 2 miles per hour or jogging at 4 miles per hour or running at 8 miles per hour, we need to consider the choices made at each hour. Let's denote the number of ways to go n miles as An. At each hour, there are three choices: walk, jog, or run. If we choose to walk, we have An-2 ways to go n miles. If we choose to jog, we have An-4 ways. If we choose to run, we have An-8 ways. Therefore, the recurrence relation is:

An = An-2 + An-4 + An-8

To find the number of ways to go 12 miles, we need to find A12. Starting from the base cases, A0 = 1, A2 = 1, A4 = 2, and A8 = 4. Using the recurrence relation, we can compute the values of A12 as follows:

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Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

given,                                  

weight of the person = 625 N

weight of the bike = 98 N        

Pressure on each Tyre = 7.60 x 10⁵ Pa

Area of contact on each Tyre = ?          

total weight of the system = 625 + 98

                                             = 723 N

Let F be the force on both the Tyre

F + F = W                                    

2 F  = 723                                    

F = 361.5 N                                

F = P A                                            

[tex]A = \dfrac{F}{P}[/tex]                          

[tex]A = \dfrac{361.5}{7.60 \times 10^5}[/tex]

A = 4.76 x 10⁻⁴ m²

The area of contact between each tire and the ground is 4.76×10⁻⁴ m².

To calculate the area of contact between each tire and the ground, we use the formula of pressure

Pressure is the force acting perpendicular per unit area to the surface of an object.

Formula:

Where:

Make A the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, The area of contact between each tire and the ground is 4.76×10⁻⁴ m².
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Answer:

Tension, T = 0.1429 N

Explanation:

Given that,

Mass of the rock, m = 0.0450 kg

Radius of the circle, r = 0.580 m

Angular speed, [tex]\omega=2.34\ rad/s[/tex]

The tension in the string is balanced by the centripetal force acting on it. It is given by :

[tex]T=\dfrac{mv^2}{r}[/tex]

Since, [tex]v=r\omega[/tex]

[tex]T=\dfrac{m(r\omega)^2}{r}[/tex]

[tex]T=\dfrac{0.0450\times (0.580\times 2.34)^2}{0.580}[/tex]

T = 0.1429 N

So, the tension in the string is 0.1429 N. Hence, this is the required solution.

To determine the percentage of the human body above the surface in the Dead Sea, compare the body's density with the water's density. Use the formula to calculate the percentage.

To determine the percentage of the human body that would be above the surface in the Dead Sea, we need to compare the density of the human body with the density of the water. The average density of the human body is 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling, whereas the density of the Dead Sea water is about 1230 kg/m3. Since the density of the human body is lower than the density of the Dead Sea water, a certain percentage of the body would be above the surface.

In the case of inhaling, the density of the human body is lower, so a larger percentage of the body would be above the surface compared to when exhaling. The exact percentage can be calculated using the formula:

Percentage above surface = (density of human body - density of water) / density of human body × 100

Once you plug in the values, you will get the percentage of the body above the Dead Sea surface.

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Answer:

The speed of the crate after the beanbag hits it is 1.2 m/s.

Explanation:

Hi there!

The momentum of the system beanbag-crate remains the same after the collision, i.e., the momentum of the system is conserved. The momentum of the system is calculated by adding the momenta of each object. So, the initial momentum (before the collision) is calculated as follows:

initial momentum = momentum of the crate + momentum of the beanbag

initial momentum = mc · vc + mb · vb

Where:

mc = mass of the crate.

vc = initial velocity of the crate.

mb = mass of the beanbag

vb = initial mass of the beanbag

With the data we have, we can calculate the initial momentum:

initial momentum  = 20 kg · 0 m/s + 1.5 kg · 10 m/s = 15 kg · m/s

Now, let´s write the equation of the momentum of the system after the collision:

final momentum = mc · vc´ + mb · vb´

Where vc´ and vb´ are the final velocity of the crate and the beanbag respectively. Let´s replace with the data we have:

final momentum = 20 kg · vc´ + 1.5 · (-6 m/s)

Since

initial momentum = final momentum

Then:

15 kg · m/s = 20 kg · vc´ + 1.5 kg · (-6 m/s)

Solving for vc´:

(15 kg · m/s + 9 kg · m/s) / 20 kg = vc´

vc´ = 1.2 m/s

The speed of the crate after the beanbag hits it is 1.2 m/s.

Using the law of conservation of momentum, the speed of the crate after the beanbag hits it is found to be 1.2 m/s in the +x direction.

The situation described can be solved using the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision. The momentum (p) of an object is its mass (m) times its velocity (v), or p = mv.

Before the collision, the 1.5-kg beanbag has a momentum of 1.5 kg * 10 m/s = 15 kg*m/s in the +x direction, and the 20-kg crate has a momentum of 0, as it is at rest. Therefore, the total momentum before the collision is 15 kg*m/s.

After the collision, the beanbag has a momentum of 1.5 kg * -6 m/s = -9 kg*m/s in the -x direction. Since the total momentum must be conserved, the crate must have a momentum of 15 kg*m/s (total initial momentum) + 9 kg*m/s (final momentum of beanbag) = 24 kg*m/s in the +x direction. Therefore, the speed (v) of the crate after the collision is its momentum divided by its mass, or v = p/m = 24 kg*m/s / 20 kg = 1.2 m/s.

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Answer:

Net filtration pressure is 2 mmHg

Solution:

As per the question:

Glomerular Pressure, [tex]P_{gh} = 46\ mmHg[/tex]

Colloidal Osmotic Pressure, [tex]P_{co} = 34\ mmHg[/tex]

Capsular Hydrostatic Pressure, [tex]P_{ch} = 10\ mmHg[/tex]

Now,

The net filtration Pressure is given by:

[tex]P_{net} = P_{gh} - (P_{co} + P_{ch})[/tex]

[tex]P_{net} = 46 - (34 + 10) = 2\ mmHg[/tex]

The net filtration pressure is 2 mmHg, which is lower than the normal value. This decrease would result in a decrease in the GFR.

The net filtration pressure is calculated by subtracting the colloid osmotic pressure and the capsular hydrostatic pressure from the glomerular hydrostatic pressure. In this case, the net filtration pressure would be 46 mmHg - 34 mmHg - 10 mmHg = 2 mmHg. This differs from the normal net filtration pressure of around 10 mmHg. A decrease in net filtration pressure would lead to a decrease in the glomerular filtration rate (GFR).

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Answer:

The separation distance between the slits is 16710.32 nm.

Explanation:

Given that,

Wavelength = 641 nm

Angle =4.4°

(a). We need to calculate the separation distance between the slits

Using formula of young's double slit

[tex]d\sin\theta=m\lambda[/tex]

[tex]d=\dfrac{m\lambda}{\sin\theta}[/tex]

Where, d = the separation distance between the slits

m = number of order

[tex]\lambda[/tex] =wavelength

Put the value into the formula

[tex]d=\dfrac{2\times641\times10^{-9}}{\sin4.4}[/tex]

[tex]d=0.00001671032\ m[/tex]

[tex]d=16710.32\ nm[/tex]

Hence, The separation distance between the slits is 16710.32 nm.

The separation distance between the slits in Young's double slit experiment can be calculated using the formula d = λx / sin(θ), where d is the separation distance, λ is the wavelength of light, θ is the angle of the fringe, and x is the distance between the fringe and the screen.

In Young's double slit experiment, the separation distance between the slits can be determined using the formula:

d = λx / sin(θ)

Where d is the separation distance between the slits, λ is the wavelength of light, θ is the angle of the fringe, and x is the distance between the fringe and the screen. In this example, the second dark fringe on either side of the central maximum is given as θ = 4.4 degrees. To calculate the separation distance between the slits, we also need to know the wavelength of the red light, which is given as λ = 641 nm.

Using the formula, we have:

d = (641 nm) * x / sin(4.4 degrees)

To find the value of d, we need to know the value of x, which is the distance between the fringe and the screen.

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Answer:

[tex] \lambda = 3.21 \ cm[/tex]

Explanation:

given,                                                                            

width of narrow slit = 4.8 cm                                

minimum angle of diffraction = θ = 42°                

wavelength of the microwave = ?                            

condition for the diffraction for single slit diffraction

      [tex]d sin \theta = m \lambda[/tex]              

for the first minima m = 1                          

      [tex]d sin \theta = \lambda[/tex]                  

wavelength of microwave radiation is equal to

      [tex] \lambda = d sin \theta[/tex]                

      [tex] \lambda = 4.8\times sin 42^0[/tex]  

      [tex] \lambda = 3.21 \ cm[/tex]                

the wavelength of microwaves is equal to [tex] \lambda = 3.21 \ cm[/tex]

The wavelength of the microwaves is approximately 0.032 m or 3.2 cm.

To find the wavelength of the microwaves given the slit width and the angle of the first diffraction minimum, you can use the diffraction formula for a single slit:

where:

Given:

Rearrange the formula to solve for the wavelength [tex]\(\lambda\)[/tex]:

Plug in the values:

First, calculate [tex]\(\sin 42^\circ\)[/tex]:

Now compute the wavelength:

So, the wavelength of the microwaves is approximately [tex]\(0.032 \text{ m}\) or \(3.2 \text{ cm}\)[/tex].

Answer:

a) 1.18 seconds

b) 8.6 m

c) 5.19 revolutions

d) 6.07 m/s

Explanation:

Step 1: Data given

radius of the ball = 11.0 cm

Initial speed of the ball = 8.50 m/s

The coefficient of kinetic friction between the ball and the lane is 0.210.

(a) For what length of time does the ball skid?

The velocity at time t can be written as v(t) = v0 + at

 ⇒ with v(t) = the velocity at time t

⇒ with v0 : the initial velocity = 8.50 m/s

⇒ with a = the acceleration (in m/s²)

   ⇒The acceleration (negative) due to friction: a = -µg

           ⇒ with µ = 0.210

          ⇒ with g = 9.81 m/s²

v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t

Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α

τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m

so α = 0.227m / 0.00484m = 46.9 rad/s²

angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t

The ball stops sliding when v(t) = ω(t) * r

8.5 - 2.06t  = 46.9*0.11*t = 5.159t

7.219t = 8.5

t = 1.18 seconds

b) How far down the lane does it skid?

s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = 8.6 m

c) How many revolutions does it make before it starts to roll?

The angular acceleration of the ball is:

α =  τ/I

 ⇒ with  τ = the torque experienced by the ball due the frictional force

   ⇒  τ = fk*R

α = fk*R /I

 ⇒ I = 2/5 m*R²

 ⇒ fk = µk*m*g

α = (µk*m*g*R)/(2/5mR²)

α = 5µk*g /2R

The angular displacement of the ball is:

∅ = 1/2αt²

⇒ The ball does not have an initial angular velocity

∅ =1/2*(5µk*g/2)*t²

∅ = 5µkgt²/4R

∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)

∅ = 32.6 rad

Number of revolutions = 32.6 rad /2π

Number of revolutions = 5.19

(d) How fast is it moving when it starts to roll?

v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = 6.07 m/s

To find the time, distance, number of revolutions, and speed at which the bowling ball starts to roll, we can use equations of motion and the relationship between linear and angular speed.

To find the answers to the given questions, we need to use the equations of linear and rotational motion as well as the relationship between linear and angular speed for the bowling ball.

(a) To find the time for the ball to skid, we can use the equation v = u + at, where v = 0 (since the ball stops skidding), u = 8.50 m/s (initial speed), and a = μg, where μ = 0.210 (coefficient of kinetic friction) and g = 9.8 m/s2 (acceleration due to gravity). Solving for t, we find t = u / (μg).

(b) To find the distance the ball skids, we can use the equation s = ut + (1/2)at2, where s is the distance skidded. Substituting the values we know, we find s = u2 / (2μg).

(c) To find the number of revolutions before the ball starts to roll, we can use the relationship between linear speed v and angular speed ω: v = Rω, where R is the radius of the ball. Solving for ω, we find ω = v / R. Since the ball is not rotating initially, the angular speed is 0. Therefore, the number of revolutions is 0.

(d) To find the speed of the ball when it starts to roll, we can use the equation v = Rω. Substituting the values we know, we find v = Rω = R(u / (Rμg)) = u / (μg).

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