A pile driver drives posts into the ground by repeatedly dropping a heavy object on them. Assume the object is dropped from the same height each time. By what factor does the energy of the pile driver-Earth system change when the mass of the object being dropped is tripled?

Answer:

a) Current in wire 1 = 0.132 A

Current in wire 2 = 0.101 A

b) Resistance of wire 1 = R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ

Resistance of wire 2 = R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ

Explanation:

Current density, J = (current) × (cross sectional area)

Current density for both wires = J = 2950 A/m²

For wire 1,

Cross sectional Area = πr² = π(0.00378²)

A₁ = 0.00004491 m²

For wire 2,

With the assumption that the strands are well banded together with no spaces in btw.

Cross sectional Area = 19 × πr² = π(0.000756)²

A₂ = 0.00003413 m²

Current in wire 1 = I₁ = J × A₁ = 2950 × 0.00004491 = 0.132 A

Current in wire 2 = I₂ = J × A₂ = 2950 × 0.00003413 = 0.101 A

b) Resistance = ρL/A

ρ = resistivity for both wires = (1.69 x 10⁻⁸) Ω.m

L = length of wire = 1.00 m for each of the two wires

A₁ = 0.00004491 m²

A₂ = 0.00003413 m²

R₁ = ρL/A₁ = (1.69 x 10⁻⁸ × 1)/0.00004491

R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ

R₂ = ρL/A₂ = (1.69 x 10⁻⁸ × 1)/0.00003413

R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ

Hope this helps!!

Answer:

[tex]v=177.95m/s[/tex]

Explanation:

First, determine circle's radius  between Earth's pole and the location. This can be calculated as:

[tex]r=R_e_a_r_t_hCos(90\frac{3}{4})\\R_e_a_r_t_h=6.37\times10^6m\\r=6.37\times10^6\times Cos67.5\textdegree\\r=2,437,693.46m\\[/tex]

The angular speed of earth is constant and is :

[tex]w=\frac{2\pi}{T}=\frac{2\pi}{24\times 3600}\\=7.3\times10^{-5}rad/s[/tex]

Velocity is:

[tex]v=wr\\=7.3\times10^{-5}\times 2,437,693.46\\v=177.95m/s[/tex]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have [tex]m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}[/tex] , where:

[tex]l[/tex] is the length

[tex]m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3[/tex]

We divide the first equation by the second equation to get:

[tex]\frac{m}{R} = \frac{d A^2}{\rho}[/tex]

[tex]A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2[/tex]

Using this Area, we find the diameter of the wire:

[tex]D = \sqrt{\frac{4A}{\pi}}[/tex]

[tex]D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}[/tex]

[tex]D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm[/tex]

To find the length, we multiply the two equations stated initially:

[tex]mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}[/tex]

[tex]l^2 = 8.534\\l = 2.92 \ m[/tex]

4.45 * 10¹⁴ J is transferred to get this mass of aluminum from 20°C to its melting point, 660⁰C.

The quantity of heat required to change the temperature of a substance is given by:

Q = mcΔT

Where Q is the heat, m is the mass of the substance, ΔT is the temperature change = final temperature - initial temperature. c is the specific heat capacity

m = 1.7 billion pounds = 77 * 10⁷ kg, ΔT = 660 - 20 = 640°C, c = 903 J/kg•K

Hence:

Q = 77 * 10⁷ kg *  903 J/kg•K * 640°C

Q = 4.45 * 10¹⁴ J

4.45 * 10¹⁴ J is transferred to get this mass of aluminum from 20°C to its melting point, 660⁰C.

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The energy required to heat 1.7 billion pounds of aluminum from 20 degrees Celsius to 660 degrees Celsius is approximately 4.398 × 10^17 Joules.

The thermal energy transferred, or heat, to raise the temperature of a substance is given by the formula q=mcΔT where 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature. For aluminum, the specific heat capacity is 0.897 Joules per gram per degree Celsius (J/g°C).

First, we need to convert 1.7 billion pounds of aluminum into grams since the specific heat value is in grams. There are about 453,592.37 grams in a pound, so this gives us about 7.711 × 10^14 grams of aluminum.

The change in temperature (ΔT) is the final temperature minus the initial temperature, or 660 degrees Celsius - 20 degrees Celsius, which equals 640 degrees Celsius.

So, to find the total energy required, we use the formula and substitute the known values: q=(7.711 × 10^14 g)*(0.897 J/g°C)*(640°C), which equals approximately 4.398 × 10^17 Joules.

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Explanation:

According to the law of conservation of energy, work done by the force is as follows.

        [tex]W_{F} = F Cos (30^{o}) \times 1.5[/tex]

                   = 64.95 J

Now, gain in potential energy is as follows.

                P.E = mgh

                      = [tex]2 \times 9.8 \times 1.5[/tex]

                      = 29.4 J

Gain in potential energy will be as follows.

           = [tex]\frac{1}{2}kx^{2}_{2} - \frac{1}{2}kx^{2}_{1}[/tex]

           = [tex]\frac{1}{2} \times 30 N/m \times [(2.5 - 1.5)^{2} - (2 - 1.5)^{2}][/tex]

           = 11.25

As,

          [tex]W_{f} = u_{1} + u_{2} + \frac{1}{2}mv^{2}[/tex]

          [tex]\frac{1}{2}mv^{2} = W_{f} - u_{1} - u_{2}[/tex]  

                   = 64.95 J - 29.4 - 11.25

                   = 24.3

              [tex]v^{2} = \frac{24.3 \times 2}{2}[/tex]

                  v = 4.92 m/s

Therefore, we can conclude that relative velocity at point B is 4.92 m/s.  

To calculate the velocity of the collar as it passes position B, we need to consider the forces acting on the collar and use Newton's second law of motion. The collar is attached to a light spring, so the force exerted by the spring can be calculated using Hooke's law. The net force acting on the collar is the sum of the force exerted by the spring and the constant 50-N force.

To calculate the velocity of the collar as it passes position B, we need to consider the forces acting on the collar and use Newton's second law of motion. The collar is attached to a light spring, so the force exerted by the spring can be calculated using Hooke's law. The net force acting on the collar is the sum of the force exerted by the spring and the constant 50-N force. We can equate this net force to the mass of the collar times its acceleration. Solving for the acceleration gives us the magnitude of the velocity as it passes position B.

Using Hooke's law, the force exerted by the spring can be calculated as:

F = k * x

where F is the force, k is the stiffness of the spring, and x is the displacement of the collar from its equilibrium position. Since the collar is attached to a light spring, we can assume that the displacement of the spring is negligible. Therefore, the force exerted by the spring is zero.

The net force is then equal to the constant 50-N force:

Net force = 50 N

Applying Newton's second law, we have:

Net force = mass * acceleration

Solving for acceleration gives us:

Acceleration = Net force / mass = 50 N / 2 kg = 25 m/s^2

Since velocity is the derivative of displacement, we can integrate the acceleration with respect to time to find the velocity:

v = a * t

where v is the velocity, a is the acceleration, and t is the time.

Since the collar is released from rest at position A, the time taken to reach position B can be determined using the equation of motion:

s = u*t + (1/2)*a*t^2

where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

Since the collar starts from rest, the initial velocity is zero. Solving for t gives us:

t = sqrt((2*s) / a) = sqrt((2*1.5 m) / 25 m/s^2) = 0.7746 s

Finally, substituting the values of acceleration and time into the equation for velocity gives us:

v = a * t = 25 m/s^2 * 0.7746 s = 19.365 m/s

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Answer:

[tex]9.198\times 10^6 m/s[/tex]

Explanation:

We are given that

Magnetic field, B=1.2 T

Radius of circular path, r=0.080 m

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

[tex]m_p=1.67\times 10^{-27} kg[/tex]

[tex]\theta=90^{\circ}[/tex]

We have to find the speed of proton.

We know that

Magnetic force, F=[tex]qvBsin\theta[/tex]

According to question

Magnetic force=Centripetal force

[tex]q_pvBsin90^{\circ}=\frac{m_pv^2}{r}[/tex]

[tex]1.6\times 10^{-19}\times 1.2=\frac{1.67\times 10^{-27}v}{0.08}[/tex]

[tex]v=\frac{1.6\times 10^{-19}\times 1.2\times 0.08}{1.67\times 10^{-27}}[/tex]

[tex]v=9.198\times 10^6 m/s[/tex]

Answer:

The overall thermal efficiency is 30%.

Explanation:

Given;

Output power = 500 kWh

input energy  per kg of coal = 6 MJ = 6 x 10⁶ J = 1.66667 kWh

1000 kg of coal will produce 1000 x 1.66667 kWh = 1666.67 kWh

Thus, total input power = 1666.67 kWh

Overall thermal efficiency = Total output power/Total input Power

Overall thermal efficiency = (500/1666.67) *100

Overall thermal efficiency = 0.29999 *100

Overall thermal efficiency = 30%

Therefore, the overall thermal efficiency is 30%.

Answer:

[tex]v(1.5)=0.7648\ m/s[/tex]

Explanation:

Dynamics

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

[tex]\displaystyle a=\frac{F}{m}[/tex]

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

[tex]F=2cos1.1t[/tex]

The variable acceleration is calculated by:

[tex]\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}[/tex]

[tex]a=0.8439cos1.1t[/tex]

The instant velocity is the integral of the acceleration:

[tex]\displaystyle v(t)=\int_{t_o}^{t_1}a.dt[/tex]

[tex]\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt[/tex]

Integrating

[tex]\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}[/tex]

[tex]\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)[/tex]

[tex]\boxed{v(1.5)=0.7648\ m/s}[/tex]

Final answer:

To find the x-component of velocity at t= 1.5 seconds for a particle under a sinusoidally varying force, we derive the equation of motion, integrate the acceleration, and calculate the velocity, resulting in [tex]v_x[/tex] = 0.567 m/s.

Explanation:

The question pertains to finding the x-component of velocity (vx) of a particle at t= 1.5 seconds, given the mass and the sinusoidally varying force. Since the force applied on the particle varies as Fx = F0cos(ωt), where F0 = 2 N and ω = 1.1 rad/s, we can find the acceleration and then integrate it with respect to time to find velocity. The acceleration ax is given by Fx/m = (2cos(1.1t))/2.37. To find the change in velocity, we integrate ax with respect to time, giving us vx = ∫ ax dt = ∫ (2cos(1.1t))/2.37 dt. Evaluating this integral from 0 to 1.5 s, we use the definite integral which simplifies to vx(t) = (2/2.37)(sin(1.1(1.5))-sin(1.1(0)))/1.1. The calculation yields vx = 0.567 m/s, indicating the particle's velocity in the x-direction at 1.5 seconds.

Answer:

The distance between the ships changing at 6PM is 21.29Km/h

Explanation:

Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h

Given

dx/dt= 35

dy/dt= 25

dv/dt= ???? at t= 6PM - 2PM= 4

Therefore t=4

We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction

So, we use:

[tex] D^2 = (150 - x)^2 + y^2 [/tex];

[tex] D^2 = (150 - 140)^2 + y^2 [/tex]

But ship B travels at t=4, at 4.25 =100 in the y-direction

so, let's use the equation:

[tex] D^2 = 10^2 + 100^2 [/tex]

[tex] = D= sqrt*(10 + 100) [/tex]

Lets use 2DD' = 2xx' + 2yy'

Differentiating with respect to t we have:

D•d(D)/dt = -(10)•dx/dt + 100•dy/dt

=100.5 d(D)/dt = (-10)•35 + (100)•25

When t=4, we have x=(140-150) =10 and y=100

[tex]= D = sqrt*(10^2 + 100^2) [/tex]

=100.5

= 100.5 dD/dt = 10.35 +100.25

= dD/dt = 21.29km/h

The distance between Ship A and Ship B is changing at approximately 21.4 km/h at 6 P.M.

Here, we use related rates. Let's define the positions of the ships -

Ship A's position at 6 P.M. :

Ship B's position at 6 P.M. :

Let’s denote the distance between the two ships at time t as D, x as the east-west distance (positive east) between Ship A and Ship B, and y as the north-south distance (positive north) from the initial position of Ship B.

Given:

We use the Pythagorean theorem to relate x, y, and D:

Differentiating both sides with respect to t:

Simplifying and solving for [tex]\frac{dD}{dt}[/tex] gives:

Plugging in the values:

Therefore, the distance between the ships is changing at approximately 21.4 km/h at 6 P.M.

Answer:

[tex]k=13\ m[/tex]

Explanation:

Given:

From the given information by the laws of reflection we can deduce the distance of the first reflection of the back of the head of person in the rear mirror.

Distance of the first reflection of the back of the head in the rear mirror from the object head:

[tex]y'=2\times y[/tex]

[tex]y'=7\ m[/tex] is the distance of the image from the object back head.

Now the total distance of this image from the front mirror:

[tex]z=y'+x[/tex]

[tex]z=7+3[/tex]

[tex]z=10\ m[/tex]

So, the total distance of the image of the back of the head in the front mirror from the person will be:

[tex]k=x+z[/tex]

[tex]k=3+10[/tex]

[tex]k=13\ m[/tex]

Final answer:

The back of the head appears to be 16.00 meters away due to multiple reflections between two parallel mirrors 6.50 meters apart, when your head is positioned 3.00 meters from the nearer mirror.

Explanation:

The question revolves around a flat mirror problem in physics where multiple reflections between two parallel mirrors are considered. When a person looks into a flat mirror, the image of any object (like the back of the head) appears to be the same distance behind the mirror as the object is in front of it. So if your head is 3.00 meters away from the nearer mirror, the first image of the back of your head will appear 3.00 meters behind that mirror.

However, since there are two mirrors, this first image will then act as an object for the second, farther mirror, creating a new image. The additional distance to this second image will be twice the distance between the two mirrors. So, the total apparent distance will be the distance to the first image plus 6.50 meters times 2, which is 16.00 meters (3.00+6.50+6.50 = 16.00 meters). Therefore, the back of your head appears to be 16.00 meters away.

Answer:

r = 0.1 Ω

Explanation:

We will use Ohm's Law in this question: V = IR, where I is the current and R is the resistance.

When the switch is open, the voltmeter reads 1.52 V, and there is no current in the circuit. We can deduce that the internal resistor in the battery causes 0.15 V to dissipate into heat, since the voltmeter reads 1.37 V when the switch is closed.

[tex]0.15 = (1.5)r\\r = 0.1~\Omega[/tex]

Answer:

Atomic Size and Mass:

convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s

Complete Question:

A volley ball is hit directly toward the ceiling in a gymnasium with a ceiling height of 10 m. If the initial vertical velocity is 13 m/s and the release height is 1.8 m will the ball hit the ceiling?

Answer:

The ball will hit the ceiling

Explanation:

Given;

Initial vertical Velocity U = 13 m/s

Height of the ceiling = 10 m

Released height of the volley ball = 1.8 m

Height traveled by the volley ball, is calculated as follows;

[tex]V^2 =U^2 -2gH[/tex]

where;

V is final vertical velocity

[tex]2gH =U^2\\\\H = \frac{U^2}{2g} = \frac{(13)^2}{2(9.8)} = 8.62 m[/tex]

Remember this ball was released from 1.8 m height and it traveled 8.62 m.

Total distance traveled = 1.8 + 8.62 = 10.42 m

Therefore, the ball will hit the ceiling

Answer: 0.321 seconds

Explanation:

Let assume that air has a temperature of 20 °C. Sound speed at given temperature is [tex]343 \frac{m}{s}[/tex]. As sound spreads at constant speed, time can be easily found by using this formula:

[tex]\Delta t = \Delta t_{far} - \Delta t_{near}[/tex]

[tex]\Delta t = \frac{x_{far}-x_{near}}{v_{sound,air}}[/tex]

[tex]\Delta{t} = \frac{120 m - 10 m}{343 \frac{m}{s} }\\\\\Delta {t} = 0.321 sec[/tex]

Answer:

[tex]I = 37.5\ A[/tex]

Explanation:

Given,

Magnetic field,B = 0.5 x 10⁻⁴ T

distance,r= 15 cm = 0.15 m

Current = ?

Using Ampere's law of magnetic field

[tex]B = \dfrac{\mu_0I}{2\pi r}[/tex]

[tex]I= \dfrac{B (2\pi r)}{\mu_0}[/tex]

[tex]I= \dfrac{0.5\times 10^{-4}\times (2\pi \times 0.15)}{4\pi \times 10^{-7}}[/tex]

[tex]I = 37.5\ A[/tex]

Current in the wire is equal to [tex]I = 37.5\ A[/tex]

The maximum current this wire can carry is equal to 37.5 Amperes.

Given the following data:

Scientific data:

In order to determine the maximum current, we would apply Ampere's law of magnetic field.

Mathematically, Ampere's law of magnetic field is given by this formula:

[tex]I=\frac{2B\pi r}{\mu_o }[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]I=\frac{2 \pi \times 0.5 \times 10^{-4}\times 0.15}{4\pi \times 10^{-7} }\\\\I=\frac{0.5 \times 10^{-4}\times 0.15}{2\pi \times 10^{-7} }[/tex]

I = 37.5 Amperes.

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The speed of the particle to pass through the Selector is 2875 m/s

Explanation:

Given -

Electric field,(We can represent as E) = 460 N/C

Magnetic field, (We can represent as B) = 0.16 T

Speed,(We can represent as  v) = ?

We know that the formula for finding the velocity ,

[tex]v = \frac{E}{B} \\\\v = \frac{460}{0.16} \\\\v = 2875m/s[/tex]

Therefore, speed of the particle to pass through the Selector is 2875 m/s

Answer: The correct option is B (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain)

Explanation:

Anterior cruciate ligament (ACL) is one of the important ligaments found at the knee joint which helps to stabilise the joint. It connects the femur to the tibia bone at the knee joint.

Anterior cruciate ligament tear is one of the common knee joint injury which is seen in individuals( especially females) involved in sports( example soccer and basketball which involves sudden change in direction causing the knee to rotate inwards)

ACL tear occurs through both contact and non contact mechanisms. The contact mechanism of ACL injury occurs when force is directly applied at the lateral part of the knee while in non contact mechanism,tear occurs when the tibia is externally rotated on the planted foot.

Research has proven that women are prone to have ACL tear than men when competing in similar sports. This disparity exists due to structural differences that pose as risk factors. These includes

- the female ACL size is smaller than the male.

- the ACL of female has a lower modulus if elasticity( that is, less stiff) than in males leading to greater joint mobility than in the male.. therefore the option, (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain) is correct.

Answer:

 v = 0.0147 m / s

Explanation:

For this exercise let's use energy conservation

Starting point. Fully stretched spring

            Em₀ = Ke = ½ k (x-x₀)²

Final point. Unstretched position

          Emf = K = ½ m v²

          Emo = Emf

         ½ k (x- x₀)² = ½ m v²

           v = √m/k    (x-x₀)

Let's calculate

            v = √(0.022 / 0.5)      (0.26-0.19)

            v = 0.0147 m / s

The speed of the mass at the mean position is 0.333 m/s

The potential energy stored in a fully stretched spring

PE = ½ kx²

where x is the stretch of the spring  = 26 -19 = 7 cm = 0.07 m

At the mean position, where x = 0, the PE stored in sprig is zero,

So according to the law of conservation of energy total energy must remain conserved so all the energy is converted into kinetic energy KE of the mass

KE = ½ mv²

where m is the mass and v is the velocity

½ kx² = ½ mv²

where k is the spring constant = 0.5 N/m

and m is the mass = 0.022 kg

[tex]v=\sqrt{\frac{k}{m} } x[/tex]

[tex]v=\sqrt{\frac{0.5}{0.022} } 0.07[/tex]

v = 0.333 m/s

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Answer:

Analyte⇒ one of analgesics

stationery phase⇒ silica

mobile phase⇒ solvent

Explanation:

during the thin layer chromatography non volatile mixtures are separated.The technique is performed on the plastic or aluminum foil that is coated with a thin layer.

Question is not complete and the missing part is;

A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.

Answer:

0.828 m/s

Explanation:

Resolving vertically, we have;

Fn and Fg act vertically. Thus,

Fn - Fg = 0 - - - - eq(1)

Resolving horizontally, we have;

Ff = ma - - - - eq(2)

Now, Fn and Fg are both mg and both will cancel out in eq 1.

Leaving us with eq 2.

So, Ff = ma

Now, Frictional force: Ff = μmg where μ is coefficient of friction.

Also, a = v²/r

Where v is linear speed or velocity

Thus,

μmg = mv²/r

m will cancel out,

Thus, μg = v²/r

Making v the subject;

rμg = v²

v = √rμg

Plugging in the relevant values,

v = √0.14 x 0.5 x 9.8

v = √0.686

v = 0.828 m/s

Final answer:

To determine the linear speed of the coin when it just begins to slip, we can use the equation frictional force = centripetal force for circular motion. By equating these two forces and solving for the linear speed, we can find the answer.

Explanation:

To determine the linear speed of the coin when it just begins to slip, we can use the equation:
frictional force = centripetal force for circular motion

The frictional force can be calculated using the equation:
frictional force = coefficient of static friction x normal force

And the centripetal force can be calculated using the equation:
centripetal force = mass of coin x acceleration towards the center of the disk

By equating these two forces and solving for the linear speed, we can find the answer.

In this case, we are given the coefficient of static friction as 0.50. We can also assume that the normal force is equal to the weight of the coin, which is the mass of the coin multiplied by the acceleration due to gravity. By plugging in these values, we can find the linear speed of the coin.

Answer:

(1)

The velocity of the center of mas of the system is= 0.801 m/s

(2)

Initial velocity of car 1 in the center of mass reference frame is =4.099 m/s

(3)

The final velocity of car 1 in the center of mass reference frame is

 - 4.099 m/s

(4)

The final velocity of car 1 in the ground (original ) reference frame is = -3.298 m/s

(5)

The final velocity of car 2 in the ground (original) reference frame is = 7.166  m/s

Explanation:

Given

m₁ = 109 kg

v₁= 4.9 m/s

m₂= 83 kg

v₂= -3.6 m/s

The two cars have an elastic collision.

(1)

The velocity of the center of mas of the system is

[tex]V_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]

       [tex]= \frac{109.4.9+83(-3.4)}{109+83}[/tex] m/s

      = 0.801 m/s

(2)

Initial velocity of car 1 in the center of mass reference frame is

[tex]V_{1,i}[/tex] = initial velocity - [tex]V_{cm}[/tex]

    = (4.9 - 0.801) m/s

    =4.099 m/s

(3)

Since the collision is elastic, the car 1 will bounce of opposite direction.

The final velocity of car 1 in the center of mass reference frame is

 [tex]V_{1,f}[/tex] = - 4.099 m/s

(4)

The final velocity of car 1 in the ground (original ) reference frame

[tex]V'_{1,f}[/tex]  =  [tex]V_{cm}+V_{1,f}[/tex]

      =(0.801- 4.099) m/s

      = - 3.298 m/s

(5)

The momentum is conserved,

[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]

[tex]\Rightarrow v'_2=\frac{m_1}{m_2}(v_1-v'_1) +v_2[/tex]  

Here [tex]v'_1= V'_{1,f}[/tex] =  - 3.298 m/s

[tex]\Rightarrow v'_2=\frac{109}{83}[4.9-(-3.298)]+(-3.6)[/tex]

      =7.166 m/s

The final velocity of car 2 in the ground (original) reference frame is = 7.166  m/s

Final answer:

The velocity of the center of mass of the bumper car system is 1.225 m/s. Car 1 has an initial velocity of 3.675 m/s in the center-of-mass reference frame. After the elastic collision, car 1's final velocity is -2.45 m/s, and car 2's final velocity is 4.825 m/s in the ground reference frame.

Explanation:

Velocity of the Center of Mass, Velocities in Reference Frames, and Final Velocities After an Elastic Collision

The velocity of the center of mass (V cm) of a system in one dimension is given by the formula:

V cm = (m1 * v1 + m2 * v2) / (m1 + m2)

In this case, we have m1 = 109 kg moving at v1 = 4.9 m/s and m2 = 83 kg moving at v2 = -3.6 m/s. The velocity of the center of mass is therefore:

V cm = (109 kg * 4.9 m/s + 83 kg * (-3.6 m/s)) / (109 kg + 83 kg)

Calculating this we get:

V cm = (534.1 kg*m/s - 298.8 kg*m/s) / 192 kg = 235.3 kg*m/s / 192 kg = 1.225 m/s (rounded to three significant figures)

The initial velocity of car 1 in the center-of-mass reference frame is given by:

u1 = v1 - V cm

Substituting the known values:

u1 = 4.9 m/s - 1.225 m/s = 3.675 m/s

In an elastic collision, velocities in the center-of-mass reference frame are mirrored. Thus, the final velocity of car 1 in the center-of-mass reference frame remains the same but in the opposite direction:

u1' = -u1 = -3.675 m/s

To find the final velocity of car 1 in the ground reference frame, we add the velocity of the center of mass:

v1' = u1' + V cm = -3.675 m/s + 1.225 m/s = -2.45 m/s

For car 2, the same principle applies. The final velocity in the center-of-mass reference frame will be the opposite of the initial, and thus:

v2' = -v2 + V cm = 3.6 m/s + 1.225 m/s = 4.825 m/s

Answer:

See the answers and the explanation below.

Explanation:

To solve this problem we must make a free body diagram with the forces applied as well as the direction. In the attached images we can see the nomenclature of the direction of the forces and the free body diagram.

a)

Sum of forces in y-axis

Fy = 140 - (220*sin(10))

Fy = 101.8 [N]

Sum of forces in x-axis

Fx = - (220*cos(10))

Fx = - 216.65 [N]

b)

For the above result, for there to be balance we realize that we need one equal to the resulting y-axis but in the opposite direction and another opposite force in direction but equal in magnitude on the x-axis.

Fy = - 101.8 [N]

Fx =   216.65 [N]

c )

Now we need to use the Pythagorean theorem to find the result of these forces.

[tex]F = \sqrt{(101.8)^{2} +(216.65)^{2} } \\F=239.4[N][/tex]

And the direction will be as follows.

α = tan^(-1) (101.8 / 216.65)

α = 25.16° (south to the east)

F = 216.65 i - 101.8 j [N]

Answer:

0.0399 m

Explanation:

We are given that

Mass of coffee=375g=[tex]\frac{375}{1000}=0.375 kg[/tex]

1kg=1000g

Depth=h=7.5 cm=[tex]7.5\times 10^{-2} m[/tex]

[tex]1 cm=10^{-2} m[/tex]

Density of coffee=[tex]\rho=1000kg/m^3[/tex]

We have to find the inside radius  of coffee mug.

We know that

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

Substitute the values

[tex]1000=\frac{0.375}{\pi r^2h}[/tex]

[tex]r^2=\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}[/tex]

By using [tex]\pi=3.14[/tex]

[tex]r=\sqrt{\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}}[/tex]

[tex]r=0.0399 m[/tex]

Hence, the inside radius=0.0399 m

Answer:

[tex]c. 5cm \times 8cm[/tex]

Explanation:

The dimensions [tex]5cm\times 8cm[/tex] have the highest cross-sectional area combination of [tex]40cm^2[/tex].

-Resistance reduces with an increase in cross sectional area.

-[tex]Reason-[/tex]Electrons have alarger area to flow through.

Answer:

[tex]v=12.65\ m.s^{-1}[/tex]

Explanation:

Given:

During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:

[tex]m.\frac{v^2}{r} =\mu.N[/tex]

where:

[tex]\mu=[/tex] coefficient of friction

[tex]N=[/tex] normal reaction force due to weight of the car

[tex]v=[/tex] velocity of the car

[tex]1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)[/tex]

[tex]v=12.65\ m.s^{-1}[/tex] is the maximum velocity at which the vehicle can turn without skidding.

Final answer:

The maximum speed at which a car can safely make a turn on a flat road, given the mass of the car, the turn's radius of curvature, and the coefficient of friction, is approximately 25 m/s. This calculation demonstrates that the car's load does not influence its ability to negotiate the turn safely on a flat surface.

Explanation:

The student is asking how to calculate the maximum speed at which a car can safely make a turn on a flat road, given the car's mass, the turn's radius of curvature, and the coefficient of friction between the tires and the road. First, let's denote the mass of the car as m, gravitational acceleration as g, the radius of curvature as R, and the coefficient of friction as μ. To find the maximum speed v, we use the fact that the centripetal force needed to make the turn must be less than or equal to the maximum static friction force, which is μmg. This gives the condition mv²/R ≤ μmg. Solving for v, we find v = √(μgR).

By plugging in the numbers: m = 1350 kg, R = 71 m, g = 9.8 m/s², and μ = 0.23, we get v = √(0.23 * 9.8 * 71) which calculates to be approximately 25 m/s. Note, because coefficients of friction are approximate, the answer is given to only two digits.

This result is quite significant as it shows that the maximum safe speed is independent of the car's mass due to the proportional relationship between friction and normal force, which in turn is proportional to mass. This implies that how heavily loaded the car is does not affect its ability to negotiate the turn, assuming a flat surface.

Answer:

8.46 N/C

Explanation:

Using Gauss law

[tex]E=\frac {kQ}{r^{2}}[/tex]

Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge

Here, K is Coulomb's constant whose value is [tex]9\times 10^{9} Nm^{2}/C^{2}[/tex]

r = 0.43 + 0.106 = 0.536 m

[tex]E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C[/tex]

The magnitude of the electric field at a point 0.106 m outside a solid metal sphere with a radius of 0.430 m and a net charge of 0.270 nC is approximately 892 N/C, to three significant figures.

The subject of your question is Physics, specifically focussing on electrostatics and the calculation of the electric field outside a charged sphere. The formula for the electric field (E) due to a point charge is given by Coulomb's Law, which is E = kQ/r^2, where 'Q' is the charge, 'r' is the distance from the charge, and 'k' is Coulomb's constant, approximately 8.99 x 10^9 N.m^2/C^2. Since the electric field due to a uniformly distributed spherical charge behaves as if all the charge is concentrated at the center, you can use this formula for the magnitude of the electric field outside the sphere.

Substituting the provided values (converted to appropriate units), we find E = (8.99 x 10^9 N.m^2/C^2 x 0.270 x 10^-9 C)/(0.536 m)^2, which gives E approximately equal to 892 N/C to three significant figures.

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Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

[tex]\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }[/tex] where [tex]V_{molar} = \frac{A}{\rho }[/tex] an and A is the atomic mass of iron.

Therefore:

[tex]\frac{2}{a^{3} } = \frac{N_{A} * p }{A}[/tex]

This implies that:

[tex]A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }[/tex]

  = [tex]\frac{4}{\sqrt{3} }r[/tex]

Assuming that there is no phase change gives:

[tex]\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }[/tex]

  = 8.60 g/m³

Answer:

Entropy Change = 0.559 Times

Explanation:

Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.

To solve this problem we will apply the concepts related to the Friction force and work. The friction force can be defined as the product between the Normal Force (Mass by gravity) and the dynamic friction constant. In the case of Work this is defined as the product of the distance traveled by the applied force. Then we will solve the points sequentially to find the answer to each point,

PART A) The friction force with the given data  is,

[tex]F_f = \mu_k mg[/tex]

Here,

[tex]\mu_k[/tex] = Kinetic coefficient

m = Mass

g = Gravitational acceleration

[tex]F_f = (0.25)(30kg)(9.8m/s^2)[/tex]

[tex]F_f = 73.5N[/tex]

PART B) The work done by the worker is the distance traveled for the previously force found, then

[tex]W_w = rF_f[/tex]

[tex]W_w = (4.5m)(73.5N)[/tex]

[tex]W_w = 330.75J[/tex]

PART C) The work done by the friction force would be the distance traveled with the previously calculated force, therefore

[tex]W_f = -rF_f[/tex]

[tex]W_f = -(4.5m)(73.5N)[/tex]

[tex]W_f = -330.75J[/tex]

[tex]W_f = -331J[/tex]

PART D) The work done by the normal force is,

[tex]W_N = N (r) Cos(90)[/tex]

[tex]W_N = 0J[/tex]

The work done by gravitational force is

[tex]W_g = rF_gcos(90)[/tex]

[tex]W_g = 0J[/tex]

PART E) The expression for the total work done is,

[tex]W_{net} = W_f +W_w +W_N+W_g[/tex]

[tex]W_{net} = 330.75-330.75+0+0[/tex]

[tex]W_{net} = 0J[/tex]

Therefore the net work done by the system is 0J

Answer:

Incomplete question

Check attachment for the diagram of the problem.

Explanation:

The acceleration of the car A is given as

a=3ft/s²

Car B is rounding a curve of radius

r=440ft

Car B is moving at constant speed of Vb=30mi/hr.

Car A reach a speed of 45mi/hr

Note, 1 mile = 5280ft

And 1 hour= 3600s

Then

Va=45mi/hr=45×5280/3600

Va=66ft/s

Also,

Vb=30mi/hour=30×5280/3600

Vb=44ft/s

Now,

a. Let write the relative velocity of car B, relative to car A

Vb = Va + Vb/a

Then,

Using triangle rule, because vectors cannot be added automatically

Vb/a²= Vb²+Va²-2Va•VbCosθ

From the given graphical question the angle between Va and Vb is 60°.

Vb/a²=44²+66² - 2•44•66Cos60

Vb/a²=1936+ 4356 - 5808Cos60

Vb/a² = 3388

Vb/a = √3388

Vb/a = 58.21 ft/s

The direction is given as

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Va/SinA = Vb/SinB = (Vb/a)/SinC

66/SinA = 44/SinB = 58.21/Sin60

Then, to get B

44/SinB = 58.21/Sin60

44Sin60/58.21  = SinB

0.6546 = SinB

B=arcsin(0.6546)

B=40.89°

b. The acceleration of Car B due to Car A.

Let write the relative acceleration  of car B, relative to car A.

Let Aa be acceleration of car A

Ab be the acceleration of car B.

Ab = Aa + Ab/a

Given the acceleration of car A

Aa=3ft/s²

Then to get the acceleration of car B, using the tangential acceleration formular

a = v²/r

Ab = Vb²/r

Ab = 44²/440

Ab = 4.4ft/s²

Using cosine rule again as above

Ab/a²= Aa²+Ab² - 2•Aa•Ab•Cosθ

Ab/a²= 3²+4.4²- 2•3•4.4•Cos30

Ab/a²= 9+19.36 - 22.863

Ab/a² = 5.497

Ab/a = √5.497

Ab/a = 2.34ft/s²

To get the direction using Sine rule again, as done above

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Aa/SinA = Ab/SinB = (Ab/a)/SinC

3/SinA = 4.4/SinB = 2.34/Sin30

Then, to get B

4.4/SinB = 2.34/Sin30

4.4Sin30/2.34 = SinB

0.9402 = SinB

B=arcsin(0.9402)

B=70.1°

Since B is obtuse, the other solution for Sine is given as

B= nπ - θ.   , when n=1

B=180-70.1

B=109.92°

To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. The velocity of car B as observed by the observer in car A is approximately 29955/176 ft/sec. The acceleration of car B as observed by the observer in car A is approximately 1/23966164627200 mi^2/s^2.

To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. Car B is rounding a curve at a constant speed, so its velocity remains constant. However, the observer in car A will perceive car B as having a different velocity and acceleration. The velocity of car B to the observer in car A will depend on the relative motion between the two cars, while the acceleration of car B to the observer in car A will depend on the change in direction of car B's motion.

Let's calculate the velocity and acceleration of car B as observed by an observer in car A:

Velocity: Since car B is rounding a curve with a radius of 440 ft and a constant speed of 30 mi/hr, we can use the formula v = rω to find the angular velocity ω. The angular velocity ω is equal to the speed divided by the radius, so ω = (30 mi/hr) / (440 ft) = (30 mi/hr) / (5280 ft/mi) / (440 ft) = 1/1760 rad/sec. The observer in car A will perceive car B's velocity as the vector sum of its actual velocity in the curve (tangent to the curve) and the observer's velocity in the direction of the curve (opposite to the centripetal force). Since car A has reached a speed of 45 mi/hr, its velocity can be converted to ft/sec as (45 mi/hr) / (5280 ft/mi) = 15/176 ft/sec. Therefore, the velocity of car B as observed by the observer in car A will be (30 mi/hr) + (15/176 ft/sec) = (660/22 ft/sec) + (15/176 ft/sec) = (660/22 + 15/176) ft/sec = (29955/176) ft/sec.

Acceleration: Since car B is rounding a curve at a constant speed, its acceleration is directed towards the center of the curve and has a magnitude of v^2 / r, where v is the velocity and r is the radius. Substituting the values, we get the acceleration as (30 mi/hr)^2 / (440 ft) = ((30 mi/hr)^2) / ((5280 ft/mi) / (440 ft)) = (900 mi^2/hr^2) / (5280 ft/mi) * (440 ft) = (900 mi^2 * ft^2) / (5280 hr^2) * (440) ft = (900 * 5280 * 440) ft^2 / hr^2 = (2119680000/5280) ft^2 / hr^2 = (400800 ft^2/hr^2) = (400800 ft^2/hr^2) * (1/3600 hr^2/s^2) * (1 mi^2 / (5280 ft)^2) = (400800 / 3600) * (1/5280)^2 mi^2/s^2 = (111/9900) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (1/266611200) mi^2/s^2 = (1/266611200) * (5280 ft/mi)^2 = (1/266611200) * 5280^2 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/19) ft^2/s^2 = (1/19) * (1/5280)^2 mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/26268580800) mi^2/s^2 = (1/26268580800) * (5280 ft/mi)^2 = (1/26268580800) * 5280^2 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/237896) ft^2/s^2 = (1/237896) * (1/5280)^2 mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/23966164627200) mi^2/s^2.

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