A 240 m long segment of wire is hanging between 2 transmission lines. What is the total magnetic force only (ignore gravitational force) on this segment of wire if the current in the wire is 500 A and the field strength is 3e-5 T

Answer:

[tex]1.6\; \rm m[/tex].

Explanation:

Let [tex]x[/tex] denote the distance (in meters) between the person and the right end of the board.

To keep the calculations simple, consider another unknown: let [tex]y[/tex] denote the support force (in Newtons) on the left end. The support force on the right end of this board would be [tex]3 \, y[/tex] (also in Newtons.)

Now there are two unknowns. At least two equations will be required for finding the exact solutions. For that, consider this board as a lever, but with two possible fulcrums. Refer to the two diagrams attached. (Not to scale.)

Calculate the torque in each situation. Note that are four external forces acting on this board at the same time. (Two support forces and two weights.) Why does each of the two diagrams show only three? In particular, why is the support force at each "fulcrum" missing? The reason is that any force acting on the lever at the fulcrum will have no direct impact on the balance between torques elsewhere on the lever. Keep in mind that the torque of each force on a lever is proportional to [tex]r[/tex], the distance between the starting point and the fulcrum. Since that missing support force starts right at the fulcrum, its [tex]r[/tex] will be zero, and it will have no torque in this context.

Hence, there are three (non-zero) torques acting on the "lever" in each diagram. For example, in the first diagram:

If the value of [tex]x[/tex] and [tex]y[/tex] are correct, these three torques should add up to zero. That is:

[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\, (8 - x))}_{\text{person}} + \underbrace{24\, y}_{\text{support}} = 0[/tex].

That gives the first equation of this system. Similarly, a different equation can be obtained using the second diagram:

[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\,x)}_{\text{person}} + \underbrace{8\, y}_{\text{support}} = 0[/tex].

Combine these two equations into a two-by-two system. Solve the system for [tex]x[/tex] and [tex]y[/tex]:

[tex]\left\lbrace\begin{aligned}&x = 1.6\\ &y = 150\end{aligned}\right.[/tex].

In other words, the person is standing at about [tex]1.6\; \rm m[/tex] from the right end of the board. The support force at the left end of the board is [tex]150\; \rm N[/tex].

Answer:

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Explanation:

Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:

[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]

Where:

[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.

[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.

[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.

After using the definition of momentum, the system is expanded:

[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]

Vectorially speaking, the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]

Where:

[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.

[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.

If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]

[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]

The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:

[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]

[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

hgccccxfcffgbbbbbbbbbbghhyhhhgdghcjyddhhyfdghhhfdgbxbbndgnncvbhcxgnjffccggshgdggjhddh

nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

Q = 20.22 x 10³ W = 20.22 KW

Answer:

the result is the quantization of __Energy__ of the particle

Explanation:

Answer:

1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec

C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km     circumference of orbit

v = C / t = 9 * 10E8 km / 3 * 10E7  sec = 30 km / sec = 18 mi/sec

Answer:

E = 2750 J at h = 5 m

Explanation:

The gravitational potential energy is given by :

[tex]E=mgh[/tex]

In this case, m is the mass of swimmer is constant at every heights. So,

At h = 1 m, E = 550 J

[tex]550=m\times 10\times 1\\\\m=55\ kg[/tex]

So, at h = 5 m, gravitational potential energy is given by :

[tex]E=55\times 10\times 5\\\\E=2750\ J[/tex]

So, the correct option is (B).

Answer:

  E = 1/9  E₀

Explanation:

In this exercise we are told that the electric field is Eo when the diameter of the balloon is D, the expression

we are asked to shorten the electric field when the diameter is 3D with the same eclectic charge

For this we can use the gauss law to find the field in the new diameter, for this we create a Gaussian surface in the form of a sphere

                Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀

In this case the lines of the electric field and the radii of the sphere are parallel, therefore the scalar product is reduced to the algebraic product and the charge inside the sphere is the initial charge Q

               A = 4π r²

               E 4π r² = Q /ε₀

               E = 1 /4πε₀     Q / r²

the value of the indicated distance is 3 times the initial diamete

                 r  = 3 D / 2

we substitute

              E = 1/4 πε₀ Q (2/ 3D)²

for the initial conditions

              E₀ = 1 / 4πε₀ Q  (2/D)²

subtitled in the equation above

         E = 1/9  E₀

Answer:

20,000 kg m/s

Explanation:

Given:

v₀ = 50 m/s

a = -5 m/s²

t = 2 s

Find: v

v = at + v₀

v = (-5 m/s²) (2 s) + (50 m/s)

v = 40 m/s

p = mv

p = (500 kg) (40 m/s)

p = 20,000 kg m/s

Answer:

2000 N

Explanation:

20 km/h = 5.56 m/s

100 km/h = 27.78 m/s

F = ma

F = m Δv/Δt

F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)

F = 2074 N

Rounded to one significant figure, the force is 2000 N.

Answer:

0.79 cm

Explanation:

The computation is shown below:-

Particle acceleration is

[tex]a = \frac{qE}{m}[/tex]

We will take d which indicates distance as from the negative plate, so the travel by proton is 0.800 cm - d at the same time

[tex]d = \frac{1}{2} a_et^2\\\\0.800 cm - d = \frac{1}{2} a_pt^2\\\\\frac{d}{0.800 cm - d} = \frac{a_e}{a_p} \\\\\frac{d}{0.800 cm - d} = \frac{m_p}{m_e} \\\\\frac{d}{0.800 cm - d} = \frac{1836m_e}{m_e}[/tex]

After solving the equation we will get 0.79 cm from the negative plate.

Therefore it is 0.79 cm far from the negative pate i.e the point at which the electron and proton pass each other

The point at which the electron and proton pass each other will be 0.79 cm.

When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.

The given data in the problem is;

d' is the distance between the two parallel plates= 0.800 cm

The acceleration is given as;

[tex]\rm a= \frac{qE}{m} \\\\[/tex]

The distance from Newton's law is found as;

[tex]d = ut+\frac{1}{2} at^2 \\\\ u=0 \\\\ d= \frac{1}{2} at^2 \\\\ d-d' = \frac{1}{2} a_pt^2 \\\\ 0.800-d= \frac{1}{2} a_pt^2 \\\\\ \frac{d}{0.800-d} =\frac{a}{a_p} \\\\ \frac{d}{0.800-d} =\frac{m_p}{m} \\\\ \frac{d}{0.800-d} =\frac{1836m_e}{m_e} \\\\ d=0.79 \ cm[/tex]

Hence the point at which the electron and proton pass each other will be 0.79 cm.

To learn more about the charge refer to the link;

https://brainly.com/question/24391667

Answer:

5000N

Explanation:

According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e

∑F = m x a             -------------(i)

From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e

∑F = F + Fₓ             -------------(ii)

Where;

Fₓ = -200N       [negative sign because the friction force opposes motion]

Combine equations(i) and (ii) together to get;

F + Fₓ = m x a

F = ma - Fₓ         -------------(iii)

Where;

m = mass of car = 1200kg

a = acceleration of the car = 4m/s²

Now substitute the values of m, a and Fₓ into equation (iii) as follows;

F = (1200 x 4) - (-200)

F = 4800 + 200

F = 5000N

Therefore, the force the thrust is providing is 5000N

Answer:

8 ft/s

Explanation:

This is a straight forward question without much ado.

It is given from the question that she walks with a speed of 8 ft/s

Answer:

T= 2p√m/k

Explanation:

This is because the period of oscillation of the mass of spring system is directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.

The period of a mass on a spring is given by the equation

T=2π√m/k.

Where T is the period,

M is mass

K is spring constant.

An increase in mass in a spring increases the period of oscillation and decrease in mass decrease period of oscillation.

When there is the relationship between the Period (T) caused by the oscillation of the mass should be considered as the T= 2p√m/k.

The mass of the spring system with respect to period of oscillation should be directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.

So the following equation should be considered

T=2π√m/k.

Here,

T is the period,

M is mass

K is spring constant.

An increase in mass in a spring rises the period of oscillation and reduce in mass decrease period of oscillation.

Learn more about mass here: https://brainly.com/question/21860379

Answer:

I hope it is correct ✌️

Answer:

The apparent weight of the passenger is 360 N

Explanation:

Given;

The mass of the passenger, m = 75 kg

radius of the loop, r = 20 m

velocity of the roller coaster, v = 10 m/s

Centripetal force acting on this passenger is given as;

[tex]F = \frac{mv^2}{r}[/tex]

where;

F is the centripetal force acting on the passenger

m is the mass of the passenger

v is the velocity of the passenger

r is the radius of the track

[tex]F = \frac{mv^2}{r} \\\\F = \frac{75*10^2}{20} \\\\F = 375 \ N[/tex]

Real weight of the passenger,

W = mg

where;

g is acceleration due to gravity

W = 75 x 9.8

W = 735 N

Apparent weight of the passenger = Real weight - Centripetal force

Apparent weight of the passenger  = 735 N - 375 N

Apparent weight of the passenger = 360 N

Therefore, the apparent weight of the passenger is 360 N

Answer:

Explanation:

Mass ( m ) = 2 kg

Speed of the object (v) = 6 metre per second

Kinetic energy =?

Now,

We have,

Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]

Plugging the values,

[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]

Reduce the numbers with Greatest Common Factor 2

[tex] = {(6)}^{2} [/tex]

Calculate

[tex] = 36 \: joule[/tex]

Hope this helps...

Good luck on your assignment...

The Kinetic energy of the object will be "36 joules".

The excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.

According to the question,

Mass of object, m = 2 kg

Speed of object, v = 6 m/s

As we know the formula,

→ Kinetic energy (K.E),

= [tex]\frac{1}{2}[/tex] × m × v²

By substituting the values, we get

= [tex]\frac{1}{2}[/tex] × 2 × (6)²

=  [tex]\frac{1}{2}[/tex] × 2 × 36

= 36 joule

Thus the above answer is appropriate.

Find out more information about Kinetic energy here:

https://brainly.com/question/25959744

Answer:

The next resonance will be 150 Hz.

Explanation:

The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.

The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.

Frecuency of the standing sound wave modes in a open-closed tube is:

fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)

The next resonance is at 90 Hz. This means that it occurs when n = 3:

f₃=3*30 Hz= 90 Hz

This means that the next resonance occurs when n = 5:

f₅=5*30 Hz= 150 Hz

The next resonance will be 150 Hz.

Answer:

Total charge = 1800C

Explanation:

Q= IT

The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom

Explanation:

Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.

Answer:

Explanation:

When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.

Along the horizontal Ux = Ucos(theta)

Along the vertical Uy = Ucos(theta)

Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.

This means that the only factor connecting horizontal and vertical components of projectile motion is its velocity since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.

Answer:

1.4 m/s

Explanation:

The minimum speed will be when the diver's initial velocity is horizontal.

First, find the time it takes for the diver to fall 10 meters.

Given:

Δy = 10 m

v₀ᵧ = 0 m/s

aᵧ = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

10 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.43 s

Now find the initial horizontal velocity.

v = (2 m) / (1.43 s)

v = 1.4 m/s

Answer:

Explanation:

Pls see diagram in attached file

Answer:

24KHz

Explanation:

See attached file

Answer:

Please see below for all the numbers to be entered in the table:

Explanation:

Coaster World:  F = 160 N; D = 40 m; T = 10 s; W = 160 * 40 = 6400 J;  V = 40/10 = 4 m/s

Wally:  F = 800 N; D = 10 m; T = 3.5 s W = 800*10 = 8000 J; V = 10/3.5 = 2.86 m/s

Elijah:  F = 1400 N; D = 800 m; T = 40 m = 2400 s; W = 1400*800 = 112000 J; V = 800/2400 = 0.33 m/s

George: F = 600 N; D = 80 m; T = 50 m = 3000 s; W = 600 * 80 = 48000 j so he should get paid: 48,000/1000= $48; V = 80/3000 = 0.027 m/s

Answer:

Down below

Explanation:

Coaster World:  F = 160 N; D = 40 m; T = 10 s; W = 160 * 40 = 6400 J;  V = 40/10 = 4 m/s

Wally:  F = 800 N; D = 10 m; T = 3.5 s W = 800*10 = 8000 J; V = 10/3.5 = 2.86 m/s

Elijah:  F = 1400 N; D = 800 m; T = 40 m = 2400 s; W = 1400*800 = 112000 J; V = 800/2400 = 0.33 m/s

George: F = 600 N; D = 80 m; T = 50 m = 3000 s; W = 600 * 80 = 48000 j so he should get paid: 48,000/1000= $48; V = 80/3000 = 0.027 m/s

Answer:

The two small charged spheres are now 4.382 cm apart

Explanation:

Given;

distance between the two small charged sphere, r = 7.59 cm

The force on each of the charged sphere can be calculated by applying Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where;

F is the force on each sphere

q₁ and q₂ are the charges of the spheres

r is the distance between the spheres

[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]

Therefore, the two small charged spheres are now 4.382 cm apart.

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

Answer:

  I = 2.18 10⁻⁴ W / m²

Explanation:

The two-slit interference pattern is described by the expression for constructive interference.

             d sin θ = m λ

If we also want to know the distribution of intensities we must perform the su of the electric field of the two waves, and find the intensity as the square of the velvet field, obtaining the expression

              I = I_max cos² ((π d /λ L) y)

where d is the separation of the slits, λ  the wavelength, L the distance to the screen e and the separation of the interference line with respect to the central maximum

let's reduce the magnitudes to the SI system

λ  = 583 nm = 583 10⁻⁹ m

L = 75.0 cm = 75.0 10⁻² m

d = 0.640 mm = 0.640 10⁻³ m

y = 0.900 mm = 0.900 10⁻³ m

let's calculate the intensity of this line

        I = 5 10⁻⁴ cos² ((π 0.640 10⁻³ /583 10⁻⁹ 0.75 10⁻²) 0.900 10⁻³)

        I = 5 10⁻⁴ cos2 (413.84)

         I = 5 10⁻⁴ 0.435

        I = 2.18 10⁻⁴ W / m²

Answer:

D is determined by distance between the telescopes.

Explanation:

Complete Question

For a human body falling through air  in  a spread edge position , the numerical value of the constant D is about [tex]D = 0.2500 kg/m[/tex]

What value of D is required to make vt = 42.7 m/s the terminal velocity of a skydiver of mass 85.0 kg . Express your answer using two significant figures?

Answer:

The value of D is   [tex]D = 0.457 \ kg/m[/tex]

Explanation:

From the question we are told that

     The terminal velocity is  [tex]v_t = 42.7 \ m/s[/tex]

     The mass of the skydiver is  [tex]m = 85.0 \ kg[/tex]

      The numerical value of  D  is  [tex]D = 0.2500 kg/m[/tex]

From the unit of D  in the question we can evaluate D as  

       [tex]D = \frac{m * g }{v^2}[/tex]

substituting values  

        [tex]D = \frac{85 * 9.8 }{(42.7)^2}[/tex]

         [tex]D = 0.457 \ kg/m[/tex]