Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answers

Answer 1

The complete question is;

In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answer:

F_top = 385.36 N

Explanation:

We are given;

mass;m = 52 kg

Time;t = 4.3 s

Diameter;d = 16m

So,Radius;r = 16/2 = 8m

The formula for the centrifugal force is given as;

F_c = mω²R

Where;

R = radius

Angular velocity;ω = 2πf

f = frequency = 1/t = 1/4.3 Hz

F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.

The force at top would be;

F_top = F_c - mg

F_top = 905.29 - (9.81 × 53) N

F_top = 385.36 N

Answer 2

The force at the top of ride will be "385.36 N".

Force and mass:

According to the question,

Rider's mass, m = 52 kg

Time, t = 4.3 s

Diameter, d = 16 m

Radius, r = [tex]\frac{16}{2}[/tex] = 8 m

Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz

We know the formula,

Centrifugal force,  [tex]F_c[/tex] = mω²R

or,

Angular velocity, ω = 2πf

By substituting the values in the above formula,

[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]

    [tex]= 905.29[/tex] N

hence,

The top force will be:

→ [tex]F_{top} = F_c[/tex] - mg

By substituting the values,

          [tex]= 905.29-(9.81\times 53)[/tex]

          [tex]= 385.36[/tex] N

Thus the above response is correct.  

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Related Questions

The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period

Answers

Answer:

Q = 23.49 C

Explanation:

We have,

Electric current drawn by the air conditioner is 16.2 A

Time, t = 1.45 s

It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,

[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]

So, the charge of 23.49 C is passing through the circuit during this period.

What is the change in internal energy of an engine if you put 15 gallon of gasoline into its tank? The energy content of gasoline is 1.5 x 106 J/gallon. All other factors, such as the engine’s temperature, are constant. How many hours the engine can work if the power of the engine’s motor is 600 W? (8 marks)

Answers

Answer:

ΔU = 2.25 x 10⁸ J

t = 104.17 s

Explanation:

The change in internal energy of the engine can be given by the following formula:

ΔU = (Mass of Gasoline)(Energy Content of Gasoline)

ΔU = (1.5 x 10⁶ J/gallon)(15 gallon)

ΔU = 2.25 x 10⁸ J

Now, for the time of operation, we use the following formula of power.

P = W/t = ΔU/t

t = ΔU/P

where,

t = time of operation = ?

ΔU = Change in internal energy = 2.25 x 10⁸ J

P = Power of motor = 600 W

Therefore,

t = (2.25 x 10⁸ J)/(600 W)

t = (375000 s)(1 h/3600 s)

t = 104.17 s

A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed

Answers

Answer:

The new angular speed is [tex]w = 6 \ rev/s[/tex]

Explanation:

From the  question we are told that

      The angular velocity of the spin is  [tex]w_o = 3 \ rev/s[/tex]

       The  original moment of inertia is  [tex]I_o[/tex]

        The new moment of inertia is  [tex]I =\frac{I_o}{2}[/tex]    

Generally angular momentum is mathematically represented as

      [tex]L = I * w[/tex]

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

         [tex]I * w = constant[/tex]

=>       [tex]I_o * w _o = I * w[/tex]

where w is the new angular speed  

  So  

          [tex]I_o * 3 = \frac{I_o}{2} * w[/tex]

=>        [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]

=>         [tex]w = 6 \ rev/s[/tex]

An amusement park ride has a vertical cylinder with an inner radius of 3.4 m, which rotates about its vertical axis. Riders stand inside against the carpeted surface and rotate with the cylinder while it accelerates to its full angular velocity. At that point the floor drops away and friction between the riders and the cylinder prevents them from sliding downward. The coefficient of static friction between the riders and the cylinder is 0.87. What minimum angular velocity in radians/second is necessary to assure that the riders will not slide down the wall?

Answers

Answer:

The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.

Explanation:

The riders will experience a centripetal force from the cylinder

[tex]F_{C}[/tex] = mrω^2    .... equ 1

where

m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of of the rider

For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as

[tex]F_{f}[/tex] = μR       ....equ 2

where

μ = coefficient of friction = 0.87

R is the normal force from the rider = mg

where

m is the rider's mass

g is the acceleration due to gravity = 9.81 m/s

substitute mg for R in equ 2, we'll have

[tex]F_{f}[/tex] = μmg     ....equ 3

Equating centripetal force of equ 1 and frictional force of equ 3, we'll get

mrω^2 = μmg

the mass of the rider cancels out, and we are left with

rω^2 = μg

ω^2 = μg/r

ω = [tex]\sqrt{\frac{ug}{r} }[/tex]

ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]

ω = 1.58 rad/second

The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s

The riders will experience a  centripetal force from the cylinder

[tex]F = mrw^2[/tex]

where  m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of the rider

For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as

f = μN

where

μ = coefficient of friction = 0.87

N is the normal force = mg

f = μmg  

Equating centripetal force of and frictional force of we'll get

[tex]mrw^2 = umg[/tex]

[tex]rw^2 = ug[/tex]

[tex]w^2 = ug/r[/tex]

[tex]w= \sqrt{ug/r}[/tex]

[tex]w= \sqrt{0.87*9.8/3.4}[/tex]  

ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.

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Which observation have scientists used to support Einstein's general theory of relativity?
The orbital path of Mercury around the Sun has changed.
O GPS clocks function at the same rate on both Earth and in space.
O The Sun has gotten more massive over time.
Objects act differently in a gravity field than in an accelerating reference frame.

Answers

Answer:

Objects act differently in a gravity field than in an accelerating reference frame.

Explanation:

The main thrust of the theory general relativity as proposed by Albert Einstein boarders on space and time as the two fundamental aspects of spacetime. Spacetime is curved in the presence of gravity, matter, energy, and momentum. The theory of general relativity explains gravity based on the way space can 'curve', that is, it seeks to relate gravitational force to the changing geometry of space-time.

The Einstein general theory of relativity has replaced Newton's ideas proposed in earlier centuries as a means of predicting gravitational interactions. This concept is quite helpful but cannot be fitted into the context of quantum mechanics due to obvious incompatibilities.

Answer:

A - The orbital path of mercury around the sun has changed.

Explanation:

got right on edg.

6. Two forces of 50 N and 30 N, respectively, are acting on an object. Find the net force (in
N) on the object if
the forces are acting in the same direction
b. the forces are acting in opposite directions.​

Answers

Answer:

same direction = 80 (n)

opposite direction = 20 (n) going one direction

Explanation:

same direction means they are added to each other

and opposite means acting on eachother

what is drift speed ? {electricity}​

Answers

Answer: In physics a drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

Explanation:

find the value of k for which the given pair of vectors are not equal
2ki +3j​ and 8i + 4kj

Answers

Answer:

5

Explanation:

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct

Answers

Complete question:

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?

A. the interior field points in a direction parallel to the exterior field

B. There is no electric field on the interior of the conducting sphere.

C. The interior field points in a direction perpendicular to the exterior field.

D. the interior field points in a direction opposite to the exterior field.

Answer:

B. There is no electric field on the interior of the conducting sphere.

Explanation:

Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).

Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of  free charges in the spherical conductor in response to any field until its neutralization.

Option B is correct.

There is no electric field on the interior of the conducting sphere.

The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.

Answers

Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  [tex]B \alpha \frac{1}{r^2}[/tex]

The magnetic field is a volume of vectors

And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.

Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .

Answers

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.

What are Velocity and Acceleration?

The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.

There are several types of velocity :

Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform Velocity

The pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.

According to the question, the given values are :

Time, t = 3.50 sec

Initial Velocity, u = 0 m/s

Use equation of motion :

v = u+at

v = 0+ g sin 18 × 3.5

v = 10.6 m/s.

So, the final velocity will be 10.6 m/s.

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Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage

Answers

Answer:

[tex]\frac{50}{\pi }[/tex]Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = [tex]\frac{50}{\pi }[/tex]Hz

Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz

Muons are elementary particles that are formed high in the atmosphere by the interactions of cosmic rays with atomic nuclei. Muons are radioactive and have average lifetimes of about two-millionths of a second. Even though they travel at almost the speed of light, they have so far to travel through the atmosphere that very few should be detected at sea level - at least according to classical physics. Laboratory measurements, however, show that muons in great number do reach the earth's surface. What is the explanation?

Answers

Answer:

Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.

Explanation:

Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.  

An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 9 cos(t) + 9 sin(t), t ≥ 0, where s is measured in centimeters and t in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time t.

Answers

Answer:

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Explanation:

Given that

s = 9 cos(t) + 9 sin(t), t ≥ 0

Then acceleration and velocity is

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

Answers

Answer:

The potential at the center of the sphere is -924 V

Explanation:

Given;

radius of the sphere, R = 0.22 m

electric field at the surface of the sphere, E = 4200 N/C

Since the electric field is directed towards the center of the sphere, the charge is negative.

The Potential is the same at every point in the sphere, and it is given as;

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)

The electric field on the sphere is also given as;

[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]

[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]

Substitute in the value of q in equation (1)

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]

Therefore, the potential at the center of the sphere is -924 V

a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric field at the 30 cm mark

Answers

Answer:

-1748*10^N/C

Explanation:

See attached file

Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.

a) Is the momentum of Cart 1 conserved?

Yes

No

It depends on M

b) Is the momentum of Cart 2 conserved?

Yes

No

It depends on M

c) Is the total momentum of Carts 1 and 2 conserved?

Yes

No

It depends on M

d) Which cart ends up moving faster?

Cart 1

Cart 2

They move at the same speed

e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?

0 m/s

4.0 m/s

5.3 m/s

16 m/s

48 m/s

64 m/s

Answers

Answer:

a) yes

b) no

c) yes

d)Cart 2 with mass [tex]\frac{M}{3}[/tex]   is expected to be more faster

e) u₂ = 48 m/s

Explanation:

a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.

b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved

c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction

note: m₁u₁ + m₂u₂ = (m₁ + m₂)v

d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.

e) Given

m₁=  M

u₁ = 16m/s

m₂ =[tex]\frac{M}{3}[/tex]

u₂ = ?

from law of conservation of momentum

m₁u₁= m₂u₂

M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)

therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)

∴u₂ = 3×16 = 48 m/s

Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and total work transfer

Answers

Answer:

Total heat transfer is positive

Total work transfer is positive

Explanation:

The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.

Heat transfer to a system is positive and that transferred from the system is negative.

Also, work done by a system is positive while the work done on the system is negative.

Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)

Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).

2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the

Answers

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

   

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

A "laser cannon" of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beam is 2E. The beam is aimed at an asteroid that is initially moving in the direction of the spacecraft. What is the acceleration of the asteroid relative to the spacecraft if the laser beam strikes the asteroid perpendicularly to its surface, and the surface is not reflecting

Answers

Answer:

Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m

Explanation:

The maximum electric field in the beam = 2E

cross-sectional area of beam = A

The intensity of an electromagnetic wave with electric field is

I = cε[tex]E_{0} ^{2}[/tex]/2

for [tex]E_{0}[/tex] = 2E

I = 2cε[tex]E^{2}[/tex]    ....equ 1

where

I is the intensity

c is the speed of light

ε is the permeability of free space

[tex]E_{0}[/tex]  is electric field

Radiation pressure of an electromagnetic wave on an absorbing surface is given as

P = I/c

substituting for I from above equ 1. we have

P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex]    ....equ 2

Also, pressure P = F/A

therefore,

F = PA    ....equ 3

where

F is the force

P is pressure

A is cross-sectional area

substitute equ 2 into equ 3, we have

F = 2ε[tex]E^{2}[/tex]A

force on a body = mass x acceleration.

that is

F = ma

therefore,

a = F/m

acceleration of the asteroid will then be

a = 2ε[tex]E^{2}[/tex]A/m

where m is the mass of the asteroid.

Consider a conducting rod of length 31 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane of the rails. At what speed (in m /s) must the sliding rod move to produce an emf of 0.75 V in a 1.75 T field?

Answers

Answer:

The speed of the rod is 1.383 m/s

Explanation:

Given;

length of the conducting rod, L = 31 cm = 0.31 m

induced emf on the rod, emf = 0.75V

magnetic field around the rod, B = 1.75 T

Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.

emef = BLv

where;

B is the magnetic field

L is length of the rod

v is the speed of the rod

v = emf / BL

v = (0.75) / (1.75 x 0.31)

v = 1.383 m/s

Therefore, the speed of the rod is 1.383 m/s

A tightly wound toroid of inner radius 1.2 cm and outer radius 2.4 cm has 960 turns of wire and carries a current of 2.5 A.

Requried:
a. What is the magnetic field at a distance of 0.9 cm from the center?
b. What is the field 1.2 cm from the center?

Answers

Answer:

a

  [tex]B = 0.0533 \ T[/tex]

b

  [tex]B = 0.04 \ T[/tex]

Explanation:

From the question we are told that

   The inner radius is [tex]r = 1.2 \ cm = 0.012 \ m[/tex]

   The  outer radius is  [tex]r_o = 2.4 \ cm = \frac{2.4}{100} = 0.024 \ m[/tex]

    The nu umber of turns is  [tex]N = 960[/tex]

    The current it is carrying is  [tex]I = 2. 5 A[/tex]

Generally the magnetic field is mathematically represented as

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

Where  [tex]\mu_o[/tex] is the permeability of free space with a constant value    

            [tex]\mu = 4\pi * 10^{-7} N/A^2[/tex]

And the given distance where the magnetic field is felt is  r =  0.9 cm  =  0.009 m

Now  substituting values

     [tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.009 }[/tex]

    [tex]B = 0.0533 \ T[/tex]

    Fro the second question the distance of the position considered from the center is  r =  1.2 cm  =  0.012 m

So the  magnetic field is  

        [tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.012 }[/tex]

        [tex]B = 0.04 \ T[/tex]

The magnetic field at a distance of 0.9 cm from the center of the toroid is 0.053 T.

The magnetic field at a distance of 1.2 cm from the center of the toroid is 0.04 T.

The given parameters;

radius of the toroid, r = 1.2 cm = 0.012 mouter radius of the toroid, R = 2.4 cm = 0.024 mnumber of turns, N = 960 turnscurrent in wire, I = 2.5 A

The magnetic field at a distance of 0.9 cm from the center of the toroid is calculated as follows;

[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.009} \\\\B = 0.053 \ T[/tex]

The magnetic field at a distance of 1.2 cm from the center of the toroid is calculated as follows;

[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.012} \\\\B = 0.04 \ T[/tex]

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can I get help please?​

Answers

Answer shown on photo

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the drift velocity of the electrons.

Answers

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = [tex]\frac{I}{nqA}[/tex]

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = [tex]\frac{I}{nqA}[/tex]

v = [tex]\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}[/tex]

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t squared plus 2 t​, where v (t )is in meters per second. ​a) How far does the particle travel during the first 2 ​sec? ​b) How far does it travel during the second 2 ​sec?

Answers

Answer:

a) 2.933 m

b) 4.534 m

Explanation:

We're given the equation

v(t) = -0.4t² + 2t

If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.

See attachment for the calculations

The conclusion of the attachment will be

7.467 - 2.933 and that is 4.534 m

Thus, The distance it travels in the second 2 sec is 4.534 m

A length of organ pipe is closed at one end. If the speed of sound is 344 m/s, what length of pipe (in cm) is needed to obtain a fundamental frequency of 50 Hz

Answers

Answer:

The length = 27.52m

Explanation:

v=f x wavelength

Calculate the ideal banking angle in degrees for a gentle turn of 1.88 km radius on a highway with a 136.3 km/hr speed limit, assuming everyone travels at the speed limit.

Answers

Answer:

Ф = 4.4°

Explanation:

given:

radius (r) = 1.88 km

velocity (v) = 136.3 km/hr

required:

banking angle ∡ ?

first:

convert 1.88 km to m = 1.88km * 1000m / 1km

r = 1880 m

convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)

v = 37.86 m/s

now.. calculate the angle

Ф = inv tan (v² / r * g)            we know that gravity = 9.8 m/s²

Ф = inv tan (37.86² / (1880 * 9.8))

Ф = 4.4°

A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about point A, that is, the point of contact between the front tire of the forklift and the ground

Answers

Answer:

The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]

Explanation:

mass of the crate = [tex]M_{C}[/tex]

speed of forklift = [tex]V_{1}[/tex]

The distance between the center of the mass and the point A = d

Recall that the angular moment is the moment of the momentum.

[tex]L = P*d[/tex]    ..... equ 1

where L is the angular momentum,

P is the momentum of the system,

d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image

Also, we know that the momentum P is the product of mass and velocity

P = mv      ....equ 2

in this case, the mass = [tex]M_{C}[/tex]

the velocity = [tex]V_{1}[/tex]

therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]

we substitute equation 2 into equation 1 to give

[tex]L = M_{C} V_{1} d[/tex]

What will be the volume and density of stone if mass of stone is 10 gram .please tell the answer fast it's very urgent I will mark as a brain me answer if you will answer it correct.

Answers

Answer:

[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]

Explanation:

Assume the stone consists of basalt, which has a density of 3.0 g/cm³.

[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]

Consider a sound wave modeled with the equation s(x, t) = 3.00 nm cos(3.50 m−1x − 1,800 s−1t). What is the maximum displacement (in nm), the wavelength (in m), the frequency (in Hz), and the speed (in m/s) of the sound wave?

Answers

Answer:

-   maximum displacement = 3.00nm

-   λ = 1.79m

-  f = 286.47 s^-1

Explanation:

You have the following equation for a sound wave:

[tex]s(x,t)=3.00nm\ cos(3.50m^{-1}x- 1,800s^{-1} t)[/tex]              (1)

The general form of the equation of a sound wave can be expressed as the following formula:

[tex]s(x,t)=Acos(kx-\omega t)[/tex]            (2)

A: amplitude of the wave = 3.00nm

k: wave number = 3.50m^-1

w: angular frequency = 1,800s^-1

- The maximum displacement of the wave is given by the amplitude of the wave, then you have:

maximum displacement = A = 3.00nm

- The wavelength is given by :

[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{3.50m^{-1}}=1.79m[/tex]

The values for the wavelength is 1.79m

- The frequency is:

[tex]f=\frac{\omega}{2\pi}=\frac{1,800s^{-1}}{2\pi}=286.47s^{-1}[/tex]

The frequency is 286.47s-1

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