A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=5.0rad/s. A second cylinder with rotational inertia I2=1.0kg·m2 rotates counterclockwise about the same axis with angular speed ω2=8.0rad/s. If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?

Answers

Answer 1

Answer:

a) 0.67 rad/sec in the clockwise direction.

b) 98.8% of the kinetic energy is lost.

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia [tex]I[/tex] = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia [tex]I[/tex] = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = [tex]I[/tex]ω

where [tex]I[/tex] is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = [tex]I[/tex]ω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = [tex]I[/tex]ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s

When they are coupled coupled together, their total rotational inertia [tex]I_{t}[/tex] = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = [tex]I_{t}[/tex][tex]w_{f}[/tex]

where [tex]I_{t}[/tex] is their total rotational inertia

[tex]w_{f}[/tex] = their final angular speed together

Final angular momentum = 3 x [tex]w_{f}[/tex] = 3[tex]w_{f}[/tex]

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 =  3[tex]w_{f}[/tex]

[tex]w_{f}[/tex] = final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s

From the first statement, the direction is clockwise

b) Rotational kinetic energy = [tex]\frac{1}{2} Iw^{2}[/tex]

where [tex]I[/tex] is the rotational inertia

[tex]w[/tex] is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = [tex]\frac{1}{2} Iw^{2}[/tex] = [tex]\frac{1}{2}*2*5^{2}[/tex] = 25 J

for second cylinder = [tex]\frac{1}{2}*1*8^{2}[/tex] = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = [tex]\frac{1}{2}I_{t}w^{2} _{f}[/tex]

where

where [tex]I_{t}[/tex] is the total rotational inertia of the cylinders

[tex]w_{f}[/tex] is final total angular speed of the coupled cylinders

Final kinetic energy =  [tex]\frac{1}{2}*3*0.67^{2}[/tex] = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = 98.8%

Answer 2

A) The angular speed of the combination of the two cylinders is; ω₃ = 0.67 rad/s

B) The percentage of the original kinetic energy lost to friction is;

percentage energy lost = 98.82%

We are given;

Rotational Inertia for first cylinder; I₁ = 2 kg.m²

Angular speed of first cylinder; ω₁ = 5 rad/s

Angular speed of second cylinder; ω₂ = 8 rad/s

Rotational Inertia for second cylinder; I₂ = 1 kg.m²

From conservation of angular momentum, we know that;

Initial angular Momentum([tex]L_{i}[/tex]) = Final angular Momentum([tex]L_{f}[/tex])

Thus;

I₁ω₁ + I₂ω₂ = I₃ω₃

Where;

ω₃ is the angular speed when the two cylinders are combined

I₃ = I₁ + I₂

I₃ = 2 + 1

I₃ = 3 kg.m²

Since the second cylinder rotates in an anticlockwise direction, then its' angular speed will be negative. Thus;

(2 * 5) + (1 * -8) = 3ω₃

10 - 8 = 3ω₃

3ω₃ = 2

ω₃ = 2/3

ω₃ = 0.67 rad/s

B) Let us find initial kinetic energy;

E_i = ¹/₂I₁ω₁² + ¹/₂I₂ω₂²

E_i = ¹/₂((2 * 5²) + (1 * 8²)

E_i = 57 J

Final kinetic energy is;

E_f = ¹/₂I₃ω₃²

E_f = ¹/₂ * 3 * 0.67²

E_f = 0.67335 J

Energy lost = 57 - 0.67335 = 56.32665 J

percentage energy lost = (56.32665/57) * 100%

percentage energy lost = 98.82%

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Related Questions

What is the current in milliamperes produced by the solar cells of a pocket calculator through which 4.2 C of charge passes in 2.7 h

Answers

Answer:

0.432mA

Explanation:

Current produced by the solar cells of the pocket calculator is expressed using the formula I = Q/t where;

Q is the charge (in Columbs)

t is the time (in seconds)

Given parameters

Q = 4.2C

t = 2.7 hrs

t = 2.7*60*60

t = 9720 seconds

Required

Current produced by the solar cell I

Substituting the given values into the formula;

I = 4.2/9720

I = 0.000432A

I = 0.432mA

Hence, the current in milliamperes produced by the solar cells of a pocket calculator is 0.432mA

of
The radii a wheel are 25 cm
and 5cm respectively, it is found
that an effort of 40N is required
to raise slowly a load 16ON
160 N. Find the Mechanical
Adventage and Effeciency,

Answers

Answer:

Explanation:

Given that

Effort = 40N

Load = 16ON

M.A = load/effort

M.A= 160N/40N

M.A = 4

Velocity ratio = V.R =radius of the wheel/radius of the axel

= 25cm/5cm

= 5

Efficiency = mechanical advantage/velocity ratio × 100/1

= 4/5 × 100/1

= 0.8×100/1

= 80%

Hence, the mechanical advantage of the machine is 4 while the efficiency is 80%.

What is the direction of the net gravitational force on the mass at the origin due to the other two masses?

Answers

Answer:

genus yds it's the

Explanation:

xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj

y=k/x, x is halved.
what happens to the value of y

Answers

Answer:

y is doubled

Explanation:

If x is halved, that means the value is doubled. Here is an exmaple:

y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.

A long, horizontal hose of diameter 5.4 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.2 cm. Water squirts from the nozzle at velocity 20 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
a) What is the velocity of the water in the hose?
b) What is the pressure differential between the water in the hose and water in the nozzle?
c) How long will it take to fill a tub of volume 120 liters with the hose?

Answers

Answer:

a) 0.988 m/s

b) 199512 Pa

c) 57.52 s

Explanation:

given that

A

A1 v1 = A2 v2

d1² v1 = d2² v2

v2 = [d1/d2]² v1

v2 = (1.2/5.4)² * 20

v2 = 0.049 * 20

v2 = 0.988 m/s

B

P + 1/2 ρ v² = K.

[p2 - p1] = 1/2 ρ [v1² - v2²]

[p2 - p1] = 1/2 * 1000 [20² - 0.988²]

[p2 - p1] = 500 * (400 - 0.976)

[p2 - p1] = 500 * 399

[p2 - p1] = 199512 Pa

C

Flow rate = AV = π [d²/ 4 ] * v

= π [0.012² / 4 ] * 20 = 0.00226 m³ /s

= π [0.054² / 4 ] * 0.988 = 0.00226 m³ /s

130 liters = 0.13 m³

t = 0.13/ 0.00226 = 57.52 s

Find the momentum of a particl with a mass of one gram moving with half the speed of light.

Answers

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

A concave mirror has a focal length of magnitude f. An object is palced in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear:______.
a) behind the mirror.
b) upright and reduced.
c) upright and enlarged.
d) inverted and reduced.
e) inverted and enlarged.

Answers

Answer:

D.

Inverted and reduced

If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

What is a concave mirror?

When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.

A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.

A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

Hence option B is correct.

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What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly

Answers

Answer:

λ = 667.85 nm

Explanation:

Let f be the frequency detected by the observer

Let v be the speed at which the observer is moving.

Now, when the direction at which the observer is moving is away from the source, we have the frequency as;

f = f_o√((1 - β)/(1 + β))

From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;

λ = λ_o√((1 + β)/(1 - β))

Where, β = Hr/c

H is hubbles constant which has a value of 0.0218 m/s • ly

c is speed of light = 3 × 10^(8) m/s

r is given as 2.40 x 10^(8) ly

Thus,

β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))

β = 0.01744

Since we are given λ_o = 656.3 nm

Then;

λ = 656.3√((1 + 0.01744)/(1 - 0.01744))

λ = 667.85 nm

As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:_______.
a. approaches zero.
b. approaches infinity.
c. approaches unity.
d. none of the above.

Answers

Answer:

b. approaches infinity

Explanation:

Because Capacitive reactance is given as Xc = 1/ωC

So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.

Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC

A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?

Answers

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      [tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

       

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is

Answers

Answer:

N2 = ¼N1

Explanation:

First of all, let's define the terms;

N1 = number of electric field lines going through the sphere of radius R

N2 = number of electric field lines going through the sphere of radius 2R

Q = the charge enclosed at the centre of concentric spheres

ε_o = a constant known as "permittivity of the free space"

E1 = Electric field in the sphere of radius R.

E2 = Electric field in the sphere of radius 2R.

A1 = Area of sphere of radius R.

A2 = Area of sphere of radius 2R

Now, from Gauss's law, the electric flux through the sphere of radius R is given by;

Φ = Q/ε_o

We also know that;

Φ = EA

Thus;

E1 × A1 = Q/ε_o

E1 = Q/(ε_o × A1)

Where A1 = 4πR²

E1 = Q/(ε_o × 4πR²)

Similarly, for the sphere of radius 2R,we have;

E2 = Q/(ε_o × 4π(2R)²)

Factorizing out to get;

E2 = ¼Q/(ε_o × 4πR²)

Comparing E2 with E1, we arrive at;

E2 = ¼E1

Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;

N2 = ¼N1

What is the density of the unknown fluid in Figure below? ρwater = 1000 kgm−3

Answers

Answer:

2500 kg/m³

Explanation:

P = P

ρgh = ρgh

ρh = ρh

(1000 kg/m³) (8.9 cm) = ρ (3.5 cm)

ρ ≈ 2500 kg/m³

A square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field B. The field is perpendicular to the plane of the loop. The loop has mass m and resistance R, and it enters the field at t = 0s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0m.

Required:
Find an expression for the loop's velocity as a function of time as it enters the magnetic field.

Answers

Answer:

v₀(1 + B²L²t/mR)

Explanation:

We know that the force on the loop is F = BIL where B = magnetic field strength, I = current and L = length of side of loop. Now the current in the loop I = ε/R where ε = induced e.m.f in the loop = BLv₀ where v₀ = velocity of loop and r = resistance of loop

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Since F = ma where a = acceleration of loop and m = mass of loop

a = F/m = B²L²v₀/mR

Using v = u + at where u = initial velocity of loop = v₀, t = time after t = 0 and v = velocity of loop after time t = 0

Substituting the value of a and u into v, we have

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

So the velocity of the loop after time t is v = v₀(1 + B²L²t/mR)

The expression for the loop's velocity as a function of time as it enters the magnetic field is v = v₀(1 + B²L²t/mR).

Calculation of the loop velocity:

As we know that

Force on the loop

F = BIL

here

B = magnetic field strength,

I = current

and L = length of side of loop.

Now

the current in the loop I = ε/R

where

ε = induced e.m.f in the loop = BLv₀

where v₀ = velocity of loop

and r = resistance of loop

So,

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Also, F = ma where a = acceleration of loop and m = mass of loop

Now

a = F/m = B²L²v₀/mR

We have to use

v = u + at

where

u = initial velocity of loop = v₀,

t = time after t = 0

and v = velocity of loop after time t = 0

So, it be like

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

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A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
(a) Is she nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct her vision?
(c) What focal length contact lens is needed, and what is its power in diopters?

Answers

Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

(c) the absolute value of the focal length (f) is given by the formula:

[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]

So it would normally be written with a negative signs in front indicating a divergent lens.

Which does not account for the fact that fish can survive the winter in ponds in temperate climate zones? 1. the density of ice versus that of water 2. the unique properties of water 3. the intermolecular bonding of water 4. the tendency for water to freeze from the bottom up

Answers

Answer:

3. the intermolecular bonding of water

Explanation:

Anomalous behavior of water is an advantage in aquatic habitat during winter. Because of some unique properties of water, it behaves irregularly. Thus, a pond or river does not freeze completely during winter.

Water has its highest density when temperature is 4[tex]^{0}C[/tex] , and lowest volume at 4[tex]^{0}C[/tex]. Thus, the denser layers of water sink accordingly until the upper layer is the least dense during winter. This layer then freeze leaving the layers below it unfrozen.

Answer:

D. The tendency for water to freeze from the bottom up.

Explanation:

Suppose you are playing hockey on a new-age ice surface for which there is no friction between the ice and the hockey puck. You wind up and hit the puck as hard as you can. After the puck loses contact with your stick, the puck will

Answers

Answer:

Not slow down or speed up.

Explanation:

Hitting the puck accelerates the speed of the puck from zero to the speed with which it leaves at the instance they lose contact. Since there is no friction between the puck and the ice, there will be no force decelerating or accelerating the hockey puck, allowing the puck to move away and remain in motion without speeding up or slowing down indefinitely theoretically.

1. What does the acronym LASER stand for? What characteristic of a laser makes it suitable for today's experiment?

Answers

Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.

Explanation:

Answer:

Light amplification by stimulated emission of radiation

A microwave oven operates at 2.4 GHz with an intensity inside the oven of 2300 W/m2 . Part A What is the amplitude of the oscillating electric field

Answers

Answer:

The amplitude of the oscillating electric field is 1316.96 N/C

Explanation:

Given;

frequency of the wave, f = 2.4 Hz

intensity of the wave, I = 2300 W/m²

Amplitude of oscillating magnetic field is given by;

[tex]B_o = \sqrt{\frac{2\mu_o I}{c} }[/tex]

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

I is intensity of wave

c is speed of light = 3 x 10⁸ m/s

[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]

The amplitude of the oscillating electric field is given by;

E₀ = cB₀

E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶

E₀ = 1316.96 N/C

Therefore, the amplitude of the oscillating electric field is 1316.96 N/C

1. A 0.430kg baseball comes off a bar and goes straight up in the air. At a height of 10.0m, the baseball has a speed of 25.3m/s. Determine the mechanical energy at the height. Show all your work. 2. What is the baseball's mechanical energy when it is at a height of 8.0m? Explain?

Answers

Answer:

180 J

Explanation:

Mechanical energy = kinetic energy + potential energy

ME = KE + PE

ME = ½ mv² + mgh

ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)

ME = 180 J

Mechanical energy is conserved, so it is 180 J at all points of the trajectory.

The baseball's mechanical energy when it is at a height of 8.0m is 180 J.

What is mechanical energy?

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time. Mechanical energy is always conserved.

Mechanical energy = kinetic energy + potential energy

Given is the mass of baseball m= 0.430 kg, height h =10m, speed v= 25.3m/s.

ME = KE + PE

ME = ½ mv² + mgh

Substitute the values, we get

ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)

ME = 180 J

Thus, the baseball's mechanical energy when it is at a height of 8.0m is 180 J.

Learn more about mechanical energy.

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A radar installation operates at 9000 MHz with an antenna (dish) that is 15 meters across. Determine the maximum distance (in kilometers) for which this system can distinguish two aircraft 100 meters apart.

Answers

Answer:

R = 36.885 km

Explanation:

In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other

The diffraction equation for slits is

            a sin θ = m λ

the first minimum occurs for m = 1

             sin θ = λ a

as the diffraction experiments the angles are very small, we approximate

             sin θ = θ

 

             θ = λ / a

This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form

            θ = 1.22 λ / a

In this problem they give us the frequency, let's find the wavelength with the relation

           c = λ f

           λ = c / f

           θ = 1.22 c/ f a

since they ask us for the distance between the planes, we can use the definition of radians

          θ = s / R

if we assume that the distance is large, we can approximate the arc to the horizontal distance

          s = x

       

we substitute

             x / R = 1.22 c / fa

             R = x f a / 1.22c

Let's reduce the magnitudes to the SI system

            f = 9000 MHz = 9 109 Hz

            a = 15 m

           x = 100 m

let's calculate

            R = 100 10⁹ 15 / (1.22 3 108)

            R = 3.6885 10⁴ m

let's reduce to km

            R = 3.6885 10¹ km

            R = 36.885 km

The hydrogen spectrum has a red line at 656 nm, and a blue line at 434 nm. What is the first order angular separation between the two spectral lines obtained with a diffraction grating with 5000 rulings/cm?

Answers

Answer:

Explanation:

grating element or slit width a = 1 x 10⁻² / 5000

= 2 x 10⁻⁶ m

angular width of first order spectral line of wavelength λ

= λ / a

for blue line angular width

= 434 x 10⁻⁹ / 2 x 10⁻⁶ radian

= 217 x 10⁻³ radian

for red line angular width

= 656 x 10⁻⁹ / 2 x 10⁻⁶ radian

= 328 x 10⁻³ radian

difference of their angular width

= 328 x 10⁻³  - 217 x 10⁻³

= 111 x 10⁻³ radian

Ans .

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current

Answers

Answer:

Explanation:

From the question we are told that

    The radius is  [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]

     The  current is  [tex]I = 4.5 \ A[/tex]

Generally the electric field is mathematically represented as

         [tex]E = \frac{J}{\sigma }[/tex]

Where [tex]\sigma[/tex] is the conductivity of  aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]

J is the current density which mathematically represented as  

      [tex]J = \frac{I}{A}[/tex]

Here A is the cross-sectional area which is mathematically represented as  

       [tex]A = \pi r^2[/tex]

       [tex]A = 3.142 * (1.4*10^{-3})^2[/tex]

       [tex]A = 6.158*10^{-6} \ m^2[/tex]

So

    [tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]

    [tex]J = 730757 A/m^2[/tex]

So

       [tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]

       [tex]E = 0.021 \ N/C[/tex]

An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the:______

Answers

Answer:

-z axis

Explanation:

According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.

a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g

Answers

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?

Answers

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism articulated to the ground at its vertex A, while vertex C is attached to the vertical cord fixed to the ground. If the coefficient of friction between the prism and the blocks is 0.4; determine the maximum angle that measures the inclination of the AC face with respect to the horizontal so that the system remains in equilibrium.

Answers

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

Convert 76.2 kilometers to meters?

Answers

Answer

76200meters

Explanation:

we know that 1km=1000meters

to convert km into meters we we divide km by meters

=76.2/1000

=76200meters

Seismic attenuation and how spherical spreading affect amplitude, can anyone explain this please!

Answers

Answer:

Hey there!

This can be a confusing topic, so it's totally fine if you get confused...

First, Seismic Attenuation is how seismic waves lose energy as they expand and spread.

Secondly, when distance increases, amplitude decreases. This is because the distance (spherical spreading would mean radius) is inversely proportional to amplitude.

Let me know if this helps :)

A steel bridge is 1000 m long at -20°C in winter. What is the change in length when the temperature rises to 40°C in summer? The average coefficient of linear expansion of this steel is 11 × 10-6 C-1.

Answers

Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature =  Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

ΔL = 0.66 m

3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.

Answers

Answer:

Option (c)

Explanation:

In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.

(C.) is the only correct statement in the list of choices.

In a series circuit, the current can flow through only one path from start to finish.

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